Class 9 Mathematics — Chapter 6: Trigonometry
Punjab Board | 2026
Exercise 6.1
Q1. Find the quadrant in which each angle lies. Write a co-terminal angle for each.
(i) 65°
Quadrant: Since \(0^\circ<65^\circ<90^\circ\), the angle lies in Quadrant I.
\[ 65^\circ+360^\circ=\boxed{425^\circ} \]
(ii) 135°
Quadrant: Since \(90^\circ<135^\circ<180^\circ\), the angle lies in Quadrant II.
\[ 135^\circ+360^\circ=\boxed{495^\circ} \]
(iii) −40°
Quadrant: The negative angle is measured clockwise, so it lies in Quadrant IV.
\[ -40^\circ+360^\circ=\boxed{320^\circ} \]
(iv) 210°
Quadrant: Since \(180^\circ<210^\circ<270^\circ\), the angle lies in Quadrant III.
\[ 210^\circ+360^\circ=\boxed{570^\circ} \]
(v) −150°
Quadrant: The angle lies in Quadrant III.
\[ -150^\circ+360^\circ=\boxed{210^\circ} \]
Q2. Convert the following angles into degrees, minutes, and seconds (DMS).
(i) 123.456°
Step 1: Degrees
The whole-number part is \(123^\circ\).
Step 2: Minutes
\[ 0.456\times60=27.36′ \]
Minutes \(=27’\)
Step 3: Seconds
\[ 0.36\times60=21.6”\approx22” \]
Answer:
\[ \boxed{123^\circ27’22”} \]
(ii) 58.7891°
Step 1: Degrees
The whole-number part is \(58^\circ\).
Step 2: Minutes
\[ 0.7891\times60=47.346′ \]
Minutes \(=47’\)
Step 3: Seconds
\[ 0.346\times60=20.76”\approx21” \]
Answer:
\[ \boxed{58^\circ47’21”} \]
(iii) 90.5678°
Step 1: Degrees
The whole-number part is \(90^\circ\).
Step 2: Minutes
\[ 0.5678\times60=34.068′ \]
Minutes \(=34’\)
Step 3: Seconds
\[ 0.068\times60=4.08”\approx4” \]
Answer:
\[ \boxed{90^\circ34’4”} \]
Q3. Convert the following angles into decimal degrees.
Formula:
\[ \text{Decimal degrees} =\text{Degrees}+\frac{\text{Minutes}}{60} +\frac{\text{Seconds}}{3600} \]
(i) \(65^\circ32’15”\)
\[ 65+\frac{32}{60}+\frac{15}{3600} \]
\[ =65+0.533333+0.004167 \]
Answer:
\[ \boxed{65.5375^\circ} \]
(ii) \(42^\circ18’45”\)
\[ 42+\frac{18}{60}+\frac{45}{3600} \]
\[ =42+0.3+0.0125 \]
Answer:
\[ \boxed{42.3125^\circ} \]
(iii) \(78^\circ45’36”\)
\[ 78+\frac{45}{60}+\frac{36}{3600} \]
\[ =78+0.75+0.01 \]
Answer:
\[ \boxed{78.76^\circ} \]
Q4. Convert the following angles into radians.
Formula:
\[ \text{Radians}=\text{Degrees}\times\frac{\pi}{180} \]
(i) \(36^\circ\)
\[ 36^\circ\times\frac{\pi}{180^\circ} =\boxed{\frac{\pi}{5}\text{ rad}} \]
(ii) \(22.5^\circ\)
\[ 22.5^\circ\times\frac{\pi}{180^\circ} =\boxed{\frac{\pi}{8}\text{ rad}} \]
(iii) \(67.5^\circ\)
\[ 67.5^\circ\times\frac{\pi}{180^\circ} =\boxed{\frac{3\pi}{8}\text{ rad}} \]
Q5. Convert the following angles into degrees.
Formula:
\[ \text{Degrees}=\text{Radians}\times\frac{180^\circ}{\pi} \]
(i) \(\frac{\pi}{16}\text{ rad}\)
\[ \frac{\pi}{16}\times\frac{180^\circ}{\pi} =\frac{180^\circ}{16} \]
Answer:
\[ \boxed{11.25^\circ} \]
(ii) \(\frac{11\pi}{5}\text{ rad}\)
\[ \frac{11\pi}{5}\times\frac{180^\circ}{\pi} =11\times36^\circ \]
Answer:
\[ \boxed{396^\circ} \]
(iii) \(\frac{7\pi}{6}\text{ rad}\)
\[ \frac{7\pi}{6}\times\frac{180^\circ}{\pi} =7\times30^\circ \]
Answer:
\[ \boxed{210^\circ} \]
Q6. Find the arc length \((\ell)\) and area \((A)\) of a sector.
Formulas:
\[ \ell=r\theta \]
\[ A=\frac12r^2\theta \]
Here, \(\theta\) is measured in radians.
(i) Given \(r=6\text{ cm}\) and \(\theta=\frac{\pi}{3}\) radians
Arc length
\[ \ell=r\theta \]
\[ =6\times\frac{\pi}{3} \]
Answer:
\[ \boxed{\ell=2\pi\approx6.28\text{ cm}} \]
Area of sector
\[ A=\frac12r^2\theta \]
\[ =\frac12\times6^2\times\frac{\pi}{3} \]
\[ =\frac12\times36\times\frac{\pi}{3} \]
Answer:
\[ \boxed{A=6\pi\approx18.85\text{ cm}^2} \]
(ii) Given \(r=\frac{4.8}{\pi}\text{ cm}\) and \(\theta=\frac{5\pi}{6}\) radians
Arc length
\[ \ell=r\theta \]
\[ =\frac{4.8}{\pi}\times\frac{5\pi}{6} \]
\[ =\frac{4.8\times5}{6} \]
Answer:
\[ \boxed{\ell=4\text{ cm}} \]
Area of sector
\[ A=\frac12r^2\theta \]
\[ =\frac12\left(\frac{4.8}{\pi}\right)^2 \times\frac{5\pi}{6} \]
\[ =\frac12\times\frac{23.04}{\pi^2} \times\frac{5\pi}{6} \]
\[ =\frac{9.6}{\pi} \]
Answer:
\[ \boxed{A\approx3.06\text{ cm}^2} \]
Q7. The central angle of a sector is \(60^\circ\) and its radius is 12 cm. Find the area of the sector and its percentage of the total circle area.
Step 1: Convert the angle into radians
\[ \theta=60^\circ\times\frac{\pi}{180^\circ} \]
\[ \theta=\frac{\pi}{3}\text{ radians} \]
Step 2: Find the area of the sector
\[ A=\frac12r^2\theta \]
\[ =\frac12\times12^2\times\frac{\pi}{3} \]
\[ =\frac12\times144\times\frac{\pi}{3} \]
\[ \boxed{A=24\pi\text{ cm}^2\approx75.40\text{ cm}^2} \]
Step 3: Find the area of the circle
\[ A=\pi r^2 \]
\[ =\pi\times12^2 \]
\[ \boxed{A=144\pi\text{ cm}^2} \]
Step 4: Find the percentage
\[ \text{Percentage} =\frac{\text{Area of sector}}{\text{Area of circle}}\times100 \]
\[ =\frac{24\pi}{144\pi}\times100 \]
\[ =\frac16\times100 \]
Answer:
\[ \boxed{16.67\%} \]
Q8. Find the percentage of the area of a sector subtending an angle of \(\frac{\pi}{8}\) radians.
Step 1: The total angle of a circle is \(2\pi\) radians.
Step 2: Find the fraction of the circle
\[ \text{Fraction}=\frac{\pi/8}{2\pi} \]
\[ =\frac1{16} \]
Step 3: Convert the fraction into a percentage
\[ \text{Percentage}=\frac1{16}\times100 \]
Answer:
\[ \boxed{6.25\%} \]
Q9. A sector with radius 12 cm and central angle \(150^\circ\) is bent to form a cone. Find its slant height and base radius.
Step 1: Find the slant height
The radius of the sector becomes the slant height of the cone.
\[ l=r=\boxed{12\text{ cm}} \]
Step 2: Find the arc length of the sector
\[ \theta=150^\circ\times\frac{\pi}{180^\circ} =\frac{5\pi}{6} \]
\[ \ell=r\theta \]
\[ =12\times\frac{5\pi}{6} \]
\[ \ell=10\pi\text{ cm} \]
Step 3: Find the base radius
The arc length of the sector becomes the circumference of the cone’s base.
\[ 2\pi r=\ell \]
\[ 2\pi r=10\pi \]
\[ r=\frac{10\pi}{2\pi} \]
Answer:
\[ \boxed{\text{Base radius}=5\text{ cm}} \]
Exercise 6.2
Q1. For the given right triangles, find the required trigonometric ratios.
Triangle (a):
Base \(=3\), Perpendicular \(=4\), Hypotenuse \(=5\)
(i) Sine
\[ \sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}} =\boxed{\frac45} \]
(ii) Cosine
\[ \cos\theta=\frac{\text{Base}}{\text{Hypotenuse}} =\boxed{\frac35} \]
(iii) Tangent
\[ \tan\theta=\frac{\text{Perpendicular}}{\text{Base}} =\boxed{\frac43} \]
(iv) Secant
\[ \sec\theta=\frac{\text{Hypotenuse}}{\text{Base}} =\boxed{\frac53} \]
(v) Cosecant
\[ \csc\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}} =\boxed{\frac54} \]
(vi) Cotangent of \(\phi\)
\[ \cot\phi=\frac{\text{Adjacent}}{\text{Opposite}} =\boxed{\frac34} \]
(vii) Tangent of \(\phi\)
\[ \tan\phi=\frac{\text{Opposite}}{\text{Adjacent}} =\boxed{\frac43} \]
(viii) Cosecant of \(\phi\)
\[ \csc\phi=\frac{\text{Hypotenuse}}{\text{Opposite}} =\boxed{\frac54} \]
(ix) Secant of \(\phi\)
\[ \sec\phi=\frac{\text{Hypotenuse}}{\text{Adjacent}} =\boxed{\frac53} \]
(x) Cosine of \(\phi\)
\[ \cos\phi=\frac{\text{Adjacent}}{\text{Hypotenuse}} =\boxed{\frac35} \]
Triangle (b):
Base \(=15\), Perpendicular \(=8\), Hypotenuse \(=17\)
\[ \sin\theta=\boxed{\frac8{17}} \]
\[ \cos\theta=\boxed{\frac{15}{17}} \]
\[ \tan\theta=\boxed{\frac8{15}} \]
Triangle (c):
Base \(=12\), Perpendicular \(=5\), Hypotenuse \(=13\)
\[ \sin\theta=\boxed{\frac5{13}} \]
\[ \cos\theta=\boxed{\frac{12}{13}} \]
\[ \tan\theta=\boxed{\frac5{12}} \]
Q2. For right triangle ABC, \(m\angle C=\theta\), \(m\angle A=\phi\), \(m\angle B=90^\circ\), and the sides are \(a,b,c\). Find the following ratios.
(i)
\[ \sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}} =\boxed{\frac cb} \]
(ii)
\[ \cos\theta=\frac{\text{Adjacent}}{\text{Hypotenuse}} =\boxed{\frac ab} \]
(iii)
\[ \tan\theta=\frac{\text{Opposite}}{\text{Adjacent}} =\boxed{\frac ca} \]
(iv)
\[ \sin\phi=\frac{\text{Opposite}}{\text{Hypotenuse}} =\boxed{\frac ab} \]
(v)
\[ \cos\phi=\frac{\text{Adjacent}}{\text{Hypotenuse}} =\boxed{\frac cb} \]
(vi)
\[ \tan\phi=\frac{\text{Opposite}}{\text{Adjacent}} =\boxed{\frac ac} \]
Q3. Verify the following identities using right triangle ABC, where \(a=3\), \(b=5\), and \(c=4\).
(i) \(\sin\theta\times\csc\theta=1\)
LHS:
\[ \frac35\times\frac53=1 \]
\[ \text{LHS}=\text{RHS} \]
Hence verified.
(ii) \(\cos\theta\times\sec\theta=1\)
LHS:
\[ \frac45\times\frac54=1 \]
\[ \text{LHS}=\text{RHS} \]
Hence verified.
(iii) \(\tan\theta\times\cot\theta=1\)
LHS:
\[ \frac34\times\frac43=1 \]
\[ \text{LHS}=\text{RHS} \]
Hence verified.
Q4. Fill in the blanks using complementary trigonometric ratios.
(i)
\[ \sin30^\circ=\sin(90^\circ-60^\circ) =\boxed{\cos60^\circ} \]
(ii)
\[ \cos30^\circ=\cos(90^\circ-60^\circ) =\boxed{\sin60^\circ} \]
(iii)
\[ \tan30^\circ=\tan(90^\circ-60^\circ) =\boxed{\cot60^\circ} \]
(iv)
\[ \tan60^\circ=\tan(90^\circ-30^\circ) =\boxed{\cot30^\circ} \]
(v)
\[ \sin60^\circ=\sin(90^\circ-30^\circ) =\boxed{\cos30^\circ} \]
(vi)
\[ \cos60^\circ=\cos(90^\circ-30^\circ) =\boxed{\sin30^\circ} \]
(vii)
\[ \sin45^\circ=\sin(90^\circ-45^\circ) =\boxed{\cos45^\circ} \]
(viii)
\[ \tan45^\circ=\tan(90^\circ-45^\circ) =\boxed{\cot45^\circ} \]
(ix)
\[ \cos45^\circ=\cos(90^\circ-45^\circ) =\boxed{\sin45^\circ} \]
Q5. In right triangle ABC, \(m\angle B=90^\circ\), \(C=60^\circ\), and \(\sin A=\frac ab\). Find the following values.
(i)
\[ \frac{m(BC)}{m(AB)}=\frac ac =\tan60^\circ=\boxed{\sqrt3} \]
(ii)
\[ \cos60^\circ=\boxed{\frac12} \]
(iii)
\[ \tan60^\circ=\boxed{\sqrt3} \]
(iv)
\[ \csc\frac{\pi}{3}=\csc60^\circ =\boxed{\frac2{\sqrt3}} \]
(v)
\[ \cot60^\circ=\boxed{\frac1{\sqrt3}} \]
(vi)
\[ \sin30^\circ=\boxed{\frac12} \]
(vii)
\[ \cos30^\circ=\boxed{\frac{\sqrt3}{2}} \]
(viii)
\[ \tan\frac{\pi}{6}=\tan30^\circ =\boxed{\frac1{\sqrt3}} \]
(ix)
\[ \sec30^\circ=\boxed{\frac2{\sqrt3}} \]
(x)
\[ \cot30^\circ=\boxed{\sqrt3} \]
Exercise 6.3
Q1. If \(\theta\) lies in the first quadrant, find the remaining trigonometric ratios.
(i) Given \(\sin\theta=\frac23\)
Perpendicular \(=2\), Hypotenuse \(=3\)
Using the Pythagorean theorem:
\[ \text{Base}=\sqrt{3^2-2^2} \]
\[ =\sqrt{9-4}=\sqrt5 \]
Remaining ratios:
\[ \cos\theta=\boxed{\frac{\sqrt5}{3}} \]
\[ \tan\theta=\boxed{\frac2{\sqrt5}} \]
\[ \csc\theta=\boxed{\frac32} \]
\[ \sec\theta=\boxed{\frac3{\sqrt5}} \]
\[ \cot\theta=\boxed{\frac{\sqrt5}{2}} \]
(ii) Given \(\cos\theta=\frac34\)
Base \(=3\), Hypotenuse \(=4\)
\[ \text{Perpendicular}=\sqrt{4^2-3^2} \]
\[ =\sqrt{16-9}=\sqrt7 \]
Remaining ratios:
\[ \sin\theta=\boxed{\frac{\sqrt7}{4}} \]
\[ \tan\theta=\boxed{\frac{\sqrt7}{3}} \]
\[ \csc\theta=\boxed{\frac4{\sqrt7}} \]
\[ \sec\theta=\boxed{\frac43} \]
\[ \cot\theta=\boxed{\frac3{\sqrt7}} \]
(iii) Given \(\tan\theta=\frac12\)
Perpendicular \(=1\), Base \(=2\)
\[ \text{Hypotenuse}=\sqrt{1^2+2^2} \]
\[ =\sqrt5 \]
Remaining ratios:
\[ \sin\theta=\boxed{\frac1{\sqrt5}} \]
\[ \cos\theta=\boxed{\frac2{\sqrt5}} \]
\[ \csc\theta=\boxed{\sqrt5} \]
\[ \sec\theta=\boxed{\frac{\sqrt5}{2}} \]
\[ \cot\theta=\boxed2 \]
(iv) Given \(\sec\theta=3\)
Hypotenuse \(=3\), Base \(=1\)
\[ \text{Perpendicular}=\sqrt{3^2-1^2} \]
\[ =\sqrt8=2\sqrt2 \]
Remaining ratios:
\[ \sin\theta=\boxed{\frac{2\sqrt2}{3}} \]
\[ \cos\theta=\boxed{\frac13} \]
\[ \tan\theta=\boxed{2\sqrt2} \]
\[ \csc\theta=\boxed{\frac3{2\sqrt2}} \]
\[ \cot\theta=\boxed{\frac1{2\sqrt2}} \]
(v) Given \(\cot\theta=\frac{\sqrt3}{2}\)
Base \(=\sqrt3\), Perpendicular \(=2\)
\[ \text{Hypotenuse}=\sqrt{(\sqrt3)^2+2^2} \]
\[ =\sqrt{3+4}=\sqrt7 \]
Remaining ratios:
\[ \sin\theta=\boxed{\frac2{\sqrt7}} \]
\[ \cos\theta=\boxed{\frac{\sqrt3}{\sqrt7}} \]
\[ \tan\theta=\boxed{\frac2{\sqrt3}} \]
\[ \csc\theta=\boxed{\frac{\sqrt7}{2}} \]
\[ \sec\theta=\boxed{\frac{\sqrt7}{\sqrt3}} \]
Q2. Prove the following trigonometric identity.
\[ (\sin\theta+\cos\theta)^2 =1+2\sin\theta\cos\theta \]
LHS:
\[ (\sin\theta+\cos\theta)^2 \]
\[ =\sin^2\theta+\cos^2\theta +2\sin\theta\cos\theta \]
Using the identity \(\sin^2\theta+\cos^2\theta=1\):
\[ =1+2\sin\theta\cos\theta \]
\[ =\text{RHS} \]
Hence proved.
Q3. Prove that \(\frac{\cos\theta}{\sin\theta}=\frac1{\tan\theta}\).
LHS:
\[ \frac{\cos\theta}{\sin\theta} =\frac1{\sin\theta/\cos\theta} \]
Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\):
\[ =\frac1{\tan\theta} \]
\[ =\text{RHS} \]
Hence proved.
Q4. Prove that \(\frac{\sin\theta}{\csc\theta}+\frac{\cos\theta}{\sec\theta}=1\).
LHS:
\[ \frac{\sin\theta}{\csc\theta} +\frac{\cos\theta}{\sec\theta} \]
\[ =\sin\theta\times\sin\theta +\cos\theta\times\cos\theta \]
\[ =\sin^2\theta+\cos^2\theta \]
\[ =1 \]
\[ =\text{RHS} \]
Hence proved.
Q5. Prove that \(\cos^2\theta-\sin^2\theta=2\cos^2\theta-1\).
LHS:
\[ \cos^2\theta-\sin^2\theta \]
Using \(\sin^2\theta=1-\cos^2\theta\):
\[ =\cos^2\theta-(1-\cos^2\theta) \]
\[ =\cos^2\theta-1+\cos^2\theta \]
\[ =2\cos^2\theta-1 \]
\[ =\text{RHS} \]
Hence proved.
Q6. Prove that \(\cos^2\theta-\sin^2\theta=1-2\sin^2\theta\).
LHS:
\[ \cos^2\theta-\sin^2\theta \]
Using \(\cos^2\theta=1-\sin^2\theta\):
\[ =(1-\sin^2\theta)-\sin^2\theta \]
\[ =1-2\sin^2\theta \]
\[ =\text{RHS} \]
Hence proved.
Q7. Prove that \(\frac{1-\sin\theta}{\cos\theta}=\frac{\cos\theta}{1+\sin\theta}\).
LHS:
\[ \frac{1-\sin\theta}{\cos\theta} \]
Multiply the numerator and denominator by \(1+\sin\theta\):
\[ =\frac{(1-\sin\theta)(1+\sin\theta)} {\cos\theta(1+\sin\theta)} \]
\[ =\frac{1-\sin^2\theta} {\cos\theta(1+\sin\theta)} \]
Using \(1-\sin^2\theta=\cos^2\theta\):
\[ =\frac{\cos^2\theta} {\cos\theta(1+\sin\theta)} \]
\[ =\frac{\cos\theta}{1+\sin\theta} \]
\[ =\text{RHS} \]
Hence proved.
Q8. Prove that \((\sec\theta-\tan\theta)^2=\frac{1-\sin\theta}{1+\sin\theta}\).
LHS:
\[ (\sec\theta-\tan\theta)^2 \]
\[ =\left(\frac1{\cos\theta} -\frac{\sin\theta}{\cos\theta}\right)^2 \]
\[ =\left(\frac{1-\sin\theta}{\cos\theta}\right)^2 \]
\[ =\frac{(1-\sin\theta)^2}{\cos^2\theta} \]
Using \(\cos^2\theta=1-\sin^2\theta\):
\[ =\frac{(1-\sin\theta)^2}{1-\sin^2\theta} \]
\[ =\frac{(1-\sin\theta)^2} {(1-\sin\theta)(1+\sin\theta)} \]
\[ =\frac{1-\sin\theta}{1+\sin\theta} \]
\[ =\text{RHS} \]
Hence proved.
Q9. Prove that \((\tan\theta+\cot\theta)^2=\sec^2\theta\csc^2\theta\).
LHS:
\[ (\tan\theta+\cot\theta)^2 \]
\[ =\left(\frac{\sin\theta}{\cos\theta} +\frac{\cos\theta}{\sin\theta}\right)^2 \]
\[ =\left(\frac{\sin^2\theta+\cos^2\theta} {\sin\theta\cos\theta}\right)^2 \]
Using \(\sin^2\theta+\cos^2\theta=1\):
\[ =\left(\frac1{\sin\theta\cos\theta}\right)^2 \]
\[ =\frac1{\sin^2\theta\cos^2\theta} \]
\[ =\sec^2\theta\csc^2\theta \]
\[ =\text{RHS} \]
Hence proved.
Q10. Prove that \(\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}=\tan\theta+\sec\theta\).
Using the identity:
\[ \sec^2\theta-\tan^2\theta=1 \]
Therefore:
\[ 1=(\sec\theta-\tan\theta)(\sec\theta+\tan\theta) \]
LHS:
\[ \frac{\tan\theta+\sec\theta-1} {\tan\theta-\sec\theta+1} \]
Replace \(1\) in the numerator:
\[ =\frac{(\tan\theta+\sec\theta) -(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)} {\tan\theta-\sec\theta+1} \]
Factor the numerator:
\[ =\frac{(\tan\theta+\sec\theta) [1-(\sec\theta-\tan\theta)]} {\tan\theta-\sec\theta+1} \]
\[ =\frac{(\tan\theta+\sec\theta) (1-\sec\theta+\tan\theta)} {\tan\theta-\sec\theta+1} \]
Since \(1-\sec\theta+\tan\theta=\tan\theta-\sec\theta+1\):
\[ =\tan\theta+\sec\theta \]
\[ =\text{RHS} \]
Hence proved.
Q11. Prove that \(\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(1+\sin\theta\cos\theta)\).
LHS:
\[ \sin^3\theta-\cos^3\theta \]
Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\):
\[ =(\sin\theta-\cos\theta) (\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta) \]
Using \(\sin^2\theta+\cos^2\theta=1\):
\[ =(\sin\theta-\cos\theta) (1+\sin\theta\cos\theta) \]
\[ =\text{RHS} \]
Hence proved.
Q12. Prove that \(\sin^6\theta-\cos^6\theta=(\sin^2\theta-\cos^2\theta)(1-\sin^2\theta\cos^2\theta)\).
LHS:
\[ \sin^6\theta-\cos^6\theta \]
\[ =(\sin^2\theta)^3-(\cos^2\theta)^3 \]
Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\):
\[ =(\sin^2\theta-\cos^2\theta) (\sin^4\theta+\sin^2\theta\cos^2\theta+\cos^4\theta) \]
Rewrite the expression in brackets:
\[ =(\sin^2\theta-\cos^2\theta) \left[(\sin^2\theta+\cos^2\theta)^2 -\sin^2\theta\cos^2\theta\right] \]
Using \(\sin^2\theta+\cos^2\theta=1\):
\[ =(\sin^2\theta-\cos^2\theta) (1-\sin^2\theta\cos^2\theta) \]
\[ =\text{RHS} \]
Hence proved.
Exercise 6.4
Q1. Find the following values without using a calculator.
(i)
\[ \sin30^\circ=\boxed{\frac12} \]
(ii)
\[ \cos30^\circ=\boxed{\frac{\sqrt3}{2}} \]
(iii)
\[ \tan\frac{\pi}{6}=\tan30^\circ =\boxed{\frac1{\sqrt3}} \]
(iv)
\[ \tan60^\circ=\boxed{\sqrt3} \]
(v)
\[ \sec60^\circ=\boxed2 \]
(vi)
\[ \cos\frac{\pi}{3}=\cos60^\circ =\boxed{\frac12} \]
(vii)
\[ \cot60^\circ=\boxed{\frac1{\sqrt3}} \]
(viii)
\[ \sin60^\circ=\boxed{\frac{\sqrt3}{2}} \]
(ix)
\[ \sec30^\circ=\boxed{\frac2{\sqrt3}} \]
(x)
\[ \csc30^\circ=\boxed2 \]
(xi)
\[ \sin45^\circ=\boxed{\frac1{\sqrt2}} \]
(xii)
\[ \cos\frac{\pi}{4}=\cos45^\circ =\boxed{\frac1{\sqrt2}} \]
Q2. Evaluate the following expressions.
(i) \(2\sin60^\circ\cos60^\circ\)
\[ =2\times\frac{\sqrt3}{2}\times\frac12 \]
Answer:
\[ \boxed{\frac{\sqrt3}{2}} \]
(ii) \(2\cos\frac{\pi}{3}\sin\frac{\pi}{3}\)
\[ =2\times\frac12\times\frac{\sqrt3}{2} \]
Answer:
\[ \boxed{\frac{\sqrt3}{2}} \]
(iii) \(2\cos45^\circ+2\sin45^\circ\)
\[ =2\times\frac1{\sqrt2}+2\times\frac1{\sqrt2} \]
\[ =\frac4{\sqrt2} \]
Answer:
\[ \boxed{2\sqrt2} \]
(iv) \(\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ\)
\[ =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2} +\frac12\times\frac12 \]
\[ =\frac34+\frac14 \]
Answer:
\[ \boxed1 \]
(v) \(\cos60^\circ\cos30^\circ-\sin60^\circ\sin30^\circ\)
\[ =\frac12\times\frac{\sqrt3}{2} -\frac{\sqrt3}{2}\times\frac12 \]
\[ =\frac{\sqrt3}{4}-\frac{\sqrt3}{4} \]
Answer:
\[ \boxed0 \]
(vi) \(\sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ\)
\[ =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2} -\frac12\times\frac12 \]
\[ =\frac34-\frac14 \]
\[ =\frac24 \]
Answer:
\[ \boxed{\frac12} \]
(vii) \(\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ\)
\[ =\frac12\times\frac{\sqrt3}{2} +\frac{\sqrt3}{2}\times\frac12 \]
\[ =\frac{\sqrt3}{4}+\frac{\sqrt3}{4} \]
Answer:
\[ \boxed{\frac{\sqrt3}{2}} \]
(viii) \(\tan\frac{\pi}{6}\cot\frac{\pi}{6}+1\)
\[ =\frac1{\sqrt3}\times\sqrt3+1 \]
\[ =1+1 \]
Answer:
\[ \boxed2 \]
Q3. If \(\sin\frac{\pi}{4}=\frac1{\sqrt2}\) and \(\cos\frac{\pi}{4}=\frac1{\sqrt2}\), find the following.
(i) \(2\sin45^\circ-2\cos45^\circ\)
\[ =2\left(\frac1{\sqrt2}\right) -2\left(\frac1{\sqrt2}\right) \]
Answer:
\[ \boxed0 \]
(ii) \(3\cos45^\circ+4\sin45^\circ\)
\[ =3\left(\frac1{\sqrt2}\right) +4\left(\frac1{\sqrt2}\right) \]
\[ =\frac7{\sqrt2} \]
Answer:
\[ \boxed{\frac7{\sqrt2}} \]
(iii) \(5\cos45^\circ-3\sin45^\circ\)
\[ =5\left(\frac1{\sqrt2}\right) -3\left(\frac1{\sqrt2}\right) \]
\[ =\frac2{\sqrt2} \]
Answer:
\[ \boxed{\sqrt2} \]
Exercise 6.5
Q1. Find the values of \(x\), \(y\), and \(z\) from the given right-angled triangles.
(i) Given \(\theta=30^\circ\) and adjacent side \(y=4\text{ cm}\)
Using \(\tan30^\circ=\frac xy\):
\[ \frac1{\sqrt3}=\frac x4 \]
\[ \boxed{x=\frac4{\sqrt3}\text{ cm}} \]
Using \(\cos30^\circ=\frac yz\):
\[ \frac{\sqrt3}{2}=\frac4z \]
\[ \boxed{z=\frac8{\sqrt3}\text{ cm}} \]
(ii) Given \(\theta=\frac{\pi}{4}=45^\circ\) and opposite side \(y=\sqrt3\text{ cm}\)
Using \(\tan45^\circ=\frac yx\):
\[ 1=\frac{\sqrt3}{x} \]
\[ \boxed{x=\sqrt3\text{ cm}} \]
Using \(\sin45^\circ=\frac yz\):
\[ \frac1{\sqrt2}=\frac{\sqrt3}{z} \]
\[ \boxed{z=\sqrt6\text{ cm}} \]
(iii) Given \(\theta=60^\circ\) and hypotenuse \(z=2\text{ cm}\)
Using \(\sin60^\circ=\frac yz\):
\[ \frac{\sqrt3}{2}=\frac y2 \]
\[ \boxed{y=\sqrt3\text{ cm}} \]
Using \(\cos60^\circ=\frac xz\):
\[ \frac12=\frac x2 \]
\[ \boxed{x=1\text{ cm}} \]
(iv) Given \(\theta=45^\circ\) and adjacent side \(y=4\text{ cm}\)
Using \(\tan45^\circ=\frac xy\):
\[ 1=\frac x4 \]
\[ \boxed{x=4\text{ cm}} \]
Using \(\cos45^\circ=\frac yz\):
\[ \frac1{\sqrt2}=\frac4z \]
\[ \boxed{z=4\sqrt2\text{ cm}} \]
Q2. Find the unknown sides and angles of the triangles.
(i) Given \(a=\sqrt3\text{ cm}\), \(c=\sqrt{13}\text{ cm}\), and \(m\angle B=90^\circ\)
Using the Pythagorean theorem:
\[ b=\sqrt{a^2+c^2} \]
\[ =\sqrt{(\sqrt3)^2+(\sqrt{13})^2} \]
\[ =\sqrt{3+13}=\sqrt{16} \]
\[ \boxed{b=4\text{ cm}} \]
Using \(\tan A=\frac ac\):
\[ \tan A=\frac{\sqrt3}{\sqrt{13}} \]
\[ \boxed{m\angle A\approx25.66^\circ} \]
Since the angles of a right triangle add up to \(180^\circ\):
\[ m\angle C=90^\circ-25.66^\circ \]
\[ \boxed{m\angle C\approx64.34^\circ} \]
(ii) Given \(a=4\text{ cm}\), \(c=4\text{ cm}\), and \(m\angle B=90^\circ\)
\[ b=\sqrt{4^2+4^2} \]
\[ =\sqrt{32} \]
\[ \boxed{b=4\sqrt2\text{ cm}} \]
Since \(a=c\), the triangle is an isosceles right triangle.
\[ \boxed{m\angle A=45^\circ,\quad m\angle C=45^\circ} \]
Q3. The side of a square field is 60 m. Find the length of its diagonal.
Using the Pythagorean theorem:
\[ d=\sqrt{60^2+60^2} \]
\[ =\sqrt{3600+3600} \]
\[ =\sqrt{7200} \]
Answer:
\[ \boxed{d=60\sqrt2\text{ m}\approx84.85\text{ m}} \]
Solve triangle ABC, where \(m\angle B=90^\circ\).
Q4. Given \(m\angle C=60^\circ\) and \(c=3\sqrt3\text{ cm}\)
\[ m\angle A=90^\circ-60^\circ \]
\[ \boxed{m\angle A=30^\circ} \]
Using \(\sin60^\circ=\frac cb\):
\[ \frac{\sqrt3}{2}=\frac{3\sqrt3}{b} \]
\[ \boxed{b=6\text{ cm}} \]
Using \(\tan60^\circ=\frac ca\):
\[ \sqrt3=\frac{3\sqrt3}{a} \]
\[ \boxed{a=3\text{ cm}} \]
Q5. Given \(m\angle C=45^\circ\) and \(a=8\text{ cm}\)
\[ m\angle A=90^\circ-45^\circ \]
\[ \boxed{m\angle A=45^\circ} \]
Since the two acute angles are equal:
\[ \boxed{c=a=8\text{ cm}} \]
Using the Pythagorean theorem:
\[ b=\sqrt{8^2+8^2} \]
\[ \boxed{b=8\sqrt2\text{ cm}} \]
Q6. Given \(a=12\text{ cm}\) and \(c=6\text{ cm}\)
\[ b=\sqrt{12^2+6^2} \]
\[ =\sqrt{144+36} \]
\[ \boxed{b=6\sqrt5\text{ cm}} \]
Using \(\tan A=\frac{12}{6}\):
\[ \tan A=2 \]
\[ \boxed{m\angle A\approx63.43^\circ} \]
\[ m\angle C=90^\circ-63.43^\circ \]
\[ \boxed{m\angle C\approx26.57^\circ} \]
Q7. Given \(m\angle A=60^\circ\) and \(c=4\text{ cm}\)
\[ m\angle C=90^\circ-60^\circ \]
\[ \boxed{m\angle C=30^\circ} \]
Using \(\tan60^\circ=\frac ac\):
\[ \sqrt3=\frac a4 \]
\[ \boxed{a=4\sqrt3\text{ cm}} \]
Using \(\cos60^\circ=\frac cb\):
\[ \frac12=\frac4b \]
\[ \boxed{b=8\text{ cm}} \]
Q8. Given \(m\angle A=60^\circ\) and \(c=4\text{ cm}\)
Using the same method as Q7:
\[ \boxed{a=4\sqrt3\text{ cm}} \]
\[ \boxed{b=8\text{ cm}} \]
\[ \boxed{m\angle C=30^\circ} \]
Q9. Given \(b=10\text{ cm}\) and \(a=6\text{ cm}\)
Using the Pythagorean theorem:
\[ c=\sqrt{10^2-6^2} \]
\[ =\sqrt{100-36} \]
\[ \boxed{c=8\text{ cm}} \]
Using \(\sin A=\frac6{10}\):
\[ \sin A=0.6 \]
\[ \boxed{m\angle A\approx36.87^\circ} \]
\[ m\angle C=90^\circ-36.87^\circ \]
\[ \boxed{m\angle C\approx53.13^\circ} \]
Q10. A canal has points Q and R on one side and point P opposite R. Given \(QR=5\) and \(QP=13\), find the width of the canal and angle PQR.
Width of the canal \((PR)\)
\[ PR=\sqrt{13^2-5^2} \]
\[ =\sqrt{169-25} \]
\[ \boxed{PR=12} \]
Angle PQR
\[ \cos(\angle PQR)=\frac5{13} \]
\[ m\angle PQR=\cos^{-1}\left(\frac5{13}\right) \]
\[ \boxed{m\angle PQR\approx67.38^\circ} \]
Q11. Calculate the length \(x\) in the adjoining figure.
Step 1: In triangle ABD
Hypotenuse \(AD=17\text{ cm}\) and \(AB=10\text{ cm}\).
\[ BD=\sqrt{17^2-10^2} \]
\[ =\sqrt{289-100} \]
\[ BD=\sqrt{189}\text{ cm} \]
Step 2: In triangle BCD
Hypotenuse \(BD=\sqrt{189}\text{ cm}\) and \(CD=8\text{ cm}\).
\[ x=\sqrt{(\sqrt{189})^2-8^2} \]
\[ =\sqrt{189-64} \]
\[ =\sqrt{125} \]
Answer:
\[ \boxed{x=5\sqrt5\text{ cm}\approx11.18\text{ cm}} \]
Q12. A ladder of length 8 m has its foot 2 m from a wall. Find the height it reaches on the wall.
Using the Pythagorean theorem:
\[ h=\sqrt{8^2-2^2} \]
\[ =\sqrt{64-4} \]
\[ =\sqrt{60} \]
Answer:
\[ \boxed{h=2\sqrt{15}\text{ m}\approx7.75\text{ m}} \]
Q13. A rectangle ABCD has diagonal \((x+9)\) m and sides \((x+7)\) m and \(x\) m. Find \(x\).
Using the Pythagorean theorem:
\[ (x+7)^2+x^2=(x+9)^2 \]
Expanding:
\[ x^2+14x+49+x^2=x^2+18x+81 \]
Simplifying:
\[ x^2-4x-32=0 \]
Factorizing:
\[ (x-8)(x+4)=0 \]
Therefore:
\[ x=8\quad\text{or}\quad x=-4 \]
Since a side length must be positive:
Answer:
\[ \boxed{x=8\text{ m}} \]
Q14. Calculate \(x\) in each case.
Case 1: Right triangle with hypotenuse 20 cm, base 12 cm, and perpendicular \(x\) cm
\[ x=\sqrt{20^2-12^2} \]
\[ =\sqrt{400-144} \]
\[ =\sqrt{256} \]
Answer:
\[ \boxed{x=16\text{ cm}} \]
Case 2: Triangle ADC is split into two right triangles at B, with \(BC=4\text{ cm}\), \(CD=5\text{ cm}\), and \(AB=BC=4\text{ cm}\)
In triangle BCD:
\[ BD=\sqrt{5^2-4^2} \]
\[ =\sqrt{25-16} \]
\[ \boxed{BD=3\text{ cm}} \]
In triangle ABD:
\[ x=\sqrt{AB^2+BD^2} \]
\[ =\sqrt{4^2+3^2} \]
\[ =\sqrt{25} \]
Answer:
\[ \boxed{x=5\text{ cm}} \]
Exercise 6.6
Q1. The angle of elevation to the top of a flagpost from a point 40 m away is \(60^\circ\). Find the height of the flagpost.
Using \(\tan60^\circ=\frac h{40}\):
\[ \sqrt3=\frac h{40} \]
\[ h=40\sqrt3 \]
Answer:
\[ \boxed{h\approx69.28\text{ m}} \]
Q2. An isosceles triangle has a vertical angle of \(120^\circ\) and a base of 10 cm. Find its altitude.
The altitude bisects the base into two equal parts:
\[ \text{Half the base}=\frac{10}{2}=5\text{ cm} \]
It also bisects the vertical angle:
\[ \text{Half the angle}=\frac{120^\circ}{2}=60^\circ \]
Using \(\tan60^\circ=\frac5h\):
\[ \sqrt3=\frac5h \]
\[ h=\frac5{\sqrt3} \]
Answer:
\[ \boxed{h\approx2.89\text{ cm}} \]
Q3. A tree is 72 m high, and the horizontal distance from the tree is 100 m. Find the angle of elevation \(\theta\).
\[ \tan\theta=\frac{72}{100} \]
\[ \tan\theta=0.72 \]
\[ \theta=\tan^{-1}(0.72) \]
Answer:
\[ \boxed{\theta\approx35.75^\circ} \]
Q4. A ladder reaches a height of 10 m while making an angle of \(60^\circ\) with the ground. Find the length of the ladder \(L\).
Using \(\sin60^\circ=\frac{10}{L}\):
\[ \frac{\sqrt3}{2}=\frac{10}{L} \]
\[ L=\frac{20}{\sqrt3} \]
Answer:
\[ \boxed{L\approx11.55\text{ m}} \]
Q5. A lighthouse is 150 m high. The angle of depression to a point is \(60^\circ\). Find the horizontal distance \(d\).
Using \(\tan60^\circ=\frac{150}{d}\):
\[ \sqrt3=\frac{150}{d} \]
\[ d=\frac{150}{\sqrt3} \]
\[ d=50\sqrt3 \]
Answer:
\[ \boxed{d\approx86.60\text{ m}} \]
Q6. The angle of elevation of a pole increases from \(15^\circ\) to \(30^\circ\) after walking 100 m towards the pole. Find the height \(h\).
Let the distance from the pole after walking be \(x\). The initial distance is \(x+100\).
Using \(\tan30^\circ=\frac hx\):
\[ \frac1{\sqrt3}=\frac hx \]
\[ x=h\sqrt3 \]
Using \(\tan15^\circ=\frac h{x+100}\):
\[ h=(x+100)\tan15^\circ \]
Substituting \(x=h\sqrt3\) and \(\tan15^\circ=2-\sqrt3\):
\[ h=(h\sqrt3+100)(2-\sqrt3) \]
Answer:
\[ \boxed{h=50\text{ m}} \]
Q7. A tower is 300 m high and casts a shadow of 450 m. Find the Sun’s angle of elevation \(\theta\).
\[ \tan\theta=\frac{300}{450} \]
\[ \tan\theta=\frac23 \]
\[ \theta=\tan^{-1}\left(\frac23\right) \]
Answer:
\[ \boxed{\theta\approx33.69^\circ} \]
Q8. The angle of elevation of a cliff changes from \(25^\circ\) to \(45^\circ\) after walking 100 m towards it. Find the height \(h\) of the cliff.
At \(45^\circ\), the horizontal distance to the cliff is \(h\).
At \(25^\circ\), the horizontal distance is \(h+100\).
Using \(\tan25^\circ=\frac h{h+100}\):
\[ 0.4663=\frac h{h+100} \]
\[ 0.4663(h+100)=h \]
\[ 0.4663h+46.63=h \]
\[ 0.5337h=46.63 \]
Answer:
\[ \boxed{h\approx87.37\text{ m}} \]
Q9. A hill is 300 m high. The angles of depression of the two shores of a river are \(70^\circ\) and \(50^\circ\). Find the width of the river and the distance to each shore.
Distance to the near shore \((d_1)\)
\[ d_1=\frac{300}{\tan70^\circ} \]
\[ d_1=\frac{300}{2.7475} \]
\[ \boxed{d_1\approx109.19\text{ m}} \]
Distance to the far shore \((d_2)\)
\[ d_2=\frac{300}{\tan50^\circ} \]
\[ d_2=\frac{300}{1.1918} \]
\[ \boxed{d_2\approx251.73\text{ m}} \]
Width of the river
\[ \text{Width}=d_2-d_1 \]
\[ =251.73-109.19 \]
Answer:
\[ \boxed{\text{Width}\approx142.54\text{ m}} \]
Q10. A kite string is 120 m long and makes an angle of elevation of \(50^\circ\). Find the height \(h\) of the kite.
Using \(\sin50^\circ=\frac h{120}\):
\[ h=120\sin50^\circ \]
\[ h=120\times0.7660 \]
Answer:
\[ \boxed{h\approx91.92\text{ m}} \]
Review Exercise 6
Q1. Multiple-Choice Questions (MCQs)
(i) The value of \(\tan^{-1}2\) in radians is:
Correct Answer: (d) 1.107 (approximately)
(ii) In a right triangle, the hypotenuse is 13 units and \(\theta=30^\circ\). Find the opposite side.
\[ \text{Opposite side}=13\sin30^\circ \]
\[ =13\times0.5 \]
Correct Answer: (a) 6.5 units
(iii) A person standing 50 m from a building sees its top at an angle of elevation of \(45^\circ\). Find the height of the building.
\[ \tan45^\circ=\frac h{50} \]
\[ 1=\frac h{50} \]
Correct Answer: (a) 50 m
(iv) \(\sec^2\theta-\tan^2\theta=?\)
Using the identity:
\[ \sec^2\theta-\tan^2\theta=1 \]
Correct Answer: (b) 1
(v) If \(\sin\theta=\frac35\), find \(\cos^2\theta\).
\[ \cos^2\theta=1-\sin^2\theta \]
\[ =1-\frac9{25} \]
\[ =\frac{16}{25} \]
Correct Answer: (c) \(\frac{16}{25}\)
(vi) Convert \(\frac{5\pi}{24}\) radians into degrees.
\[ \frac{5\pi}{24}\times\frac{180^\circ}{\pi} =37.5^\circ \]
Correct Answer: (b) \(37.5^\circ\)
(vii) Convert \(292.5^\circ\) into radians.
\[ 292.5^\circ\times\frac{\pi}{180^\circ} =\frac{13\pi}{8} \]
Correct Answer: (b) \(\frac{13\pi}{8}\text{ rad}\)
(viii) Select the valid identity.
Correct Answer: (a) \(\cos\left(\frac\pi2-\theta\right)=\sin\theta\)
(ix) Find \(\sin60^\circ\).
Correct Answer: (d) \(\frac{\sqrt3}{2}\)
(x) Find \(\cos2100\pi+\sin2100\pi\).
Since \(\cos(2100\pi)=1\) and \(\sin(2100\pi)=0\):
\[ \cos(2100\pi)+\sin(2100\pi)=1+0=1 \]
Correct Answer: (a) 1
Q2. Angle Conversions
(a) Convert degrees into radians in terms of \(\pi\).
(i) \(255^\circ\)
\[ 255^\circ\times\frac{\pi}{180^\circ} =\frac{17\pi}{12} \]
Answer:
\[ \boxed{\frac{17\pi}{12}\text{ rad}} \]
(ii) \(75^\circ45’\)
Convert minutes into decimal degrees:
\[ 75^\circ45’=75.75^\circ \]
\[ 75.75\times\frac{\pi}{180} =\frac{101\pi}{240} \]
Answer:
\[ \boxed{\frac{101\pi}{240}\text{ rad}} \]
(iii) \(142.5^\circ\)
\[ 142.5^\circ\times\frac{\pi}{180^\circ} \]
Answer:
\[ \boxed{\frac{19\pi}{24}\text{ rad}} \]
(b) Convert radians into degrees.
(i) \(\frac{17\pi}{24}\text{ rad}\)
\[ \frac{17\pi}{24}\times\frac{180^\circ}{\pi} =127.5^\circ \]
Answer:
\[ \boxed{127^\circ30′} \]
(ii) \(\frac{7\pi}{12}\text{ rad}\)
\[ \frac{7\pi}{12}\times\frac{180^\circ}{\pi} \]
Answer:
\[ \boxed{105^\circ} \]
(iii) \(\frac{11\pi}{16}\text{ rad}\)
\[ \frac{11\pi}{16}\times\frac{180^\circ}{\pi} =123.75^\circ \]
Answer:
\[ \boxed{123^\circ45′} \]
Q3. Prove the following identities.
(i) \(\frac{\sin\theta}{1-\cos\theta}=\frac{1+\cos\theta}{\sin\theta}\)
LHS:
\[ \frac{\sin\theta}{1-\cos\theta} \]
Multiply the numerator and denominator by \(1+\cos\theta\):
\[ =\frac{\sin\theta(1+\cos\theta)} {(1-\cos\theta)(1+\cos\theta)} \]
\[ =\frac{\sin\theta(1+\cos\theta)} {1-\cos^2\theta} \]
Using \(1-\cos^2\theta=\sin^2\theta\):
\[ =\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta} \]
\[ =\frac{1+\cos\theta}{\sin\theta} \]
\[ =\text{RHS} \]
Hence proved.
(ii) \(\sin\theta(\csc\theta-\sin\theta)=\frac1{\sec^2\theta}\)
LHS:
\[ \sin\theta(\csc\theta-\sin\theta) \]
\[ =\sin\theta\times\csc\theta-\sin^2\theta \]
\[ =1-\sin^2\theta \]
\[ =\cos^2\theta \]
Since \(\cos^2\theta=\frac1{\sec^2\theta}\):
\[ =\frac1{\sec^2\theta} \]
\[ =\text{RHS} \]
Hence proved.
(iii) \(\frac{\csc\theta-\sec\theta}{\csc\theta+\sec\theta}=\frac{1-\tan\theta}{1+\tan\theta}\)
LHS:
\[ \frac{\csc\theta-\sec\theta}{\csc\theta+\sec\theta} \]
\[ =\frac{\frac1{\sin\theta}-\frac1{\cos\theta}} {\frac1{\sin\theta}+\frac1{\cos\theta}} \]
\[ =\frac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta} \]
Divide the numerator and denominator by \(\cos\theta\):
\[ =\frac{1-\tan\theta}{1+\tan\theta} \]
\[ =\text{RHS} \]
Hence proved.
(iv) \(\tan\theta+\cot\theta=\frac1{\sin\theta\cos\theta}\)
LHS:
\[ \tan\theta+\cot\theta \]
\[ =\frac{\sin\theta}{\cos\theta} +\frac{\cos\theta}{\sin\theta} \]
\[ =\frac{\sin^2\theta+\cos^2\theta} {\sin\theta\cos\theta} \]
Using \(\sin^2\theta+\cos^2\theta=1\):
\[ =\frac1{\sin\theta\cos\theta} \]
\[ =\text{RHS} \]
Hence proved.
Q4. If \(\tan\theta=\frac3{\sqrt2}\) and \(\theta\) lies in Quadrant I, find the remaining trigonometric ratios.
Perpendicular \(=3\), Base \(=\sqrt2\)
Using the Pythagorean theorem:
\[ \text{Hypotenuse}=\sqrt{3^2+(\sqrt2)^2} \]
\[ =\sqrt{9+2} \]
\[ =\sqrt{11} \]
Remaining ratios:
\[ \sin\theta=\boxed{\frac3{\sqrt{11}}} \]
\[ \cos\theta=\boxed{\frac{\sqrt2}{\sqrt{11}}} \]
\[ \csc\theta=\boxed{\frac{\sqrt{11}}3} \]
\[ \sec\theta=\boxed{\frac{\sqrt{11}}{\sqrt2}} \]
\[ \cot\theta=\boxed{\frac{\sqrt2}{3}} \]
Q5. A point on the ground has an angle of elevation of \(28^\circ\) to the top of a 30 m building. Find the horizontal distance \(d\).
Using \(\tan28^\circ=\frac{30}{d}\):
\[ d=\frac{30}{\tan28^\circ} \]
\[ d=\frac{30}{0.5317} \]
Answer:
\[ \boxed{d\approx56.42\text{ m}} \]
Q6. A ladder is 10 m long and makes an angle of \(65^\circ\) with the ground. Find the height \(h\) it reaches on the wall.
Using \(\sin65^\circ=\frac h{10}\):
\[ h=10\sin65^\circ \]
\[ h=10\times0.9063 \]
Answer:
\[ \boxed{h\approx9.06\text{ m}} \]
Notes by Sir Dilawar
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