Mathematics · Introduction to Trigonometry · NCERT Exercises

Exercise 8.3

Complete, independently verified solutions for NCERT Exercise 8.3.

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Mathematics · Chapter 8 · NCERT Exercises

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Question 1

identities · in terms of cot

Express sin A, sec A and tan A in terms of cot A.

Ratios in terms of cot AA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Ratios in terms of cot A — explanatory sketch (not to scale).

Solution

Board-exam working:

1+cot⁡2A=cosec⁡2A1+\cot^2A=\cosec^2A
cosec⁡A=1+cot⁡2A\cosec A=\sqrt{1+\cot^2A}
sin⁡A=1cosec⁡A=11+cot⁡2A\sin A=\frac1{\cosec A}=\frac1{\sqrt{1+\cot^2A}}
tan⁡A=1cot⁡A\tan A=\frac1{\cot A}
sec⁡2A=1+tan⁡2A=1+1cot⁡2A\sec^2A=1+\tan^2A=1+\frac1{\cot^2A}
sec⁡A=1+cot⁡2Acot⁡A\sec A=\frac{\sqrt{1+\cot^2A}}{\cot A}

Final answer

sin⁡A=11+cot⁡2A,sec⁡A=1+cot⁡2Acot⁡A,tan⁡A=1cot⁡A\sin A=\frac1{\sqrt{1+\cot^2A}},\quad \sec A=\frac{\sqrt{1+\cot^2A}}{\cot A},\quad \tan A=\frac1{\cot A}

Question

Question 2

identities · in terms of sec

Write all other trigonometric ratios of angle A in terms of sec A.

Ratios in terms of sec AA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Ratios in terms of sec A — explanatory sketch (not to scale).

Solution

Board-exam working:

cos⁡A=1sec⁡A\cos A=\frac1{\sec A}
sec⁡2A=1+tan⁡2A⇒tan⁡A=sec⁡2A−1\sec^2A=1+\tan^2A\Rightarrow\tan A=\sqrt{\sec^2A-1}
cot⁡A=1sec⁡2A−1\cot A=\frac1{\sqrt{\sec^2A-1}}
sin⁡A=tan⁡Acos⁡A=sec⁡2A−1sec⁡A\sin A=\tan A\cos A=\frac{\sqrt{\sec^2A-1}}{\sec A}
cosec⁡A=sec⁡Asec⁡2A−1\cosec A=\frac{\sec A}{\sqrt{\sec^2A-1}}

Final answer

cos⁡A=1sec⁡A,tan⁡A=sec⁡2A−1,cot⁡A=1sec⁡2A−1,sin⁡A=sec⁡2A−1sec⁡A,cosec⁡A=sec⁡Asec⁡2A−1\cos A=\frac1{\sec A},\quad \tan A=\sqrt{\sec^2A-1},\quad \cot A=\frac1{\sqrt{\sec^2A-1}},\quad \sin A=\frac{\sqrt{\sec^2A-1}}{\sec A},\quad \cosec A=\frac{\sec A}{\sqrt{\sec^2A-1}}

Question

Question 3

multiple choice · identities

Choose and justify: (i) 9sec²A − 9tan²A; (ii) (1 + tanθ + secθ)(1 + cotθ − cosecθ); (iii) (sec A + tan A)(1 − sin A); (iv) (1 + tan²A)/(1 + cot²A).

Reciprocal ratios and identitiesA right triangle marks angle θ, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseθABC
Reciprocal ratios and identities — explanatory sketch (not to scale).

Solution

Board-exam working:

(i)

9(sec⁡2A−tan⁡2A)=99(\sec^2A-\tan^2A)=9

Option B.

(ii) Let t = tan θ, so cot θ = 1/t and cosec θ = sec θ/t.

(1+t+sec⁡θ)(1+1t−sec⁡θt)(1+t+\sec\theta)\left(1+\frac1t-\frac{\sec\theta}{t}\right)
=(1+t+sec⁡θ)(1+t−sec⁡θ)t=\frac{(1+t+\sec\theta)(1+t-\sec\theta)}t
=(1+t)2−sec⁡2θt=1+2t+t2−(1+t2)t=2=\frac{(1+t)^2-\sec^2\theta}{t}=\frac{1+2t+t^2-(1+t^2)}t=2

Option C.

(iii)

(sec⁡A+tan⁡A)(1−sin⁡A)=1+sin⁡Acos⁡A(1−sin⁡A)(\sec A+\tan A)(1-\sin A)=\frac{1+\sin A}{\cos A}(1-\sin A)
=1−sin⁡2Acos⁡A=cos⁡A=\frac{1-\sin^2A}{\cos A}=\cos A

Option D.

(iv)

1+tan⁡2A1+cot⁡2A=sec⁡2Acosec⁡2A=tan⁡2A\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A

Option D.

Final answer

(i) B; (ii) C; (iii) D; (iv) D.

Question

Question 4

prove identities · ten subparts

Prove the following identities for acute angles where the expressions are defined:

(i) (cosec⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ\text{(i) }(\cosec\theta-\cot\theta)^2=\frac{1-\cos\theta}{1+\cos\theta}
(ii) cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\text{(ii) }\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A
(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcosec⁡θ\text{(iii) }\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\cosec\theta
(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\text{(iv) }\frac{1+\sec A}{\sec A}=\frac{\sin^2A}{1-\cos A}
(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec⁡A+cot⁡A\text{(v) }\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cosec A+\cot A
(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\text{(vi) }\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A
(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\text{(vii) }\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\tan\theta
(viii) (sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A\text{(viii) }(\sin A+\cosec A)^2+(\cos A+\sec A)^2=7+\tan^2A+\cot^2A
(ix) (cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\text{(ix) }(\cosec A-\sin A)(\sec A-\cos A)=\frac1{\tan A+\cot A}
(x) 1+tan⁡2A1+cot⁡2A=(1−tan⁡A1−cot⁡A)2=tan⁡2A\text{(x) }\frac{1+\tan^2A}{1+\cot^2A}=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan^2A
Pythagorean identities for acute anglesA right triangle marks angle θ, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseθABC
Pythagorean identities for acute angles — explanatory sketch (not to scale).

Solution

Board-exam working:

(i)

(cosec⁡θ−cot⁡θ)2=(1−cos⁡θsin⁡θ)2(\cosec\theta-\cot\theta)^2=\left(\frac{1-\cos\theta}{\sin\theta}\right)^2
=(1−cos⁡θ)21−cos⁡2θ=1−cos⁡θ1+cos⁡θ=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}=\frac{1-\cos\theta}{1+\cos\theta}

(ii)

cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}
=1−sin⁡Acos⁡A+1+sin⁡Acos⁡A=2cos⁡A=2sec⁡A=\frac{1-\sin A}{\cos A}+\frac{1+\sin A}{\cos A}=\frac2{\cos A}=2\sec A

(iii)

tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}
=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)−cos⁡2θsin⁡θ(sin⁡θ−cos⁡θ)=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}
=sin⁡3θ−cos⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ)=\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}
=1+sin⁡θcos⁡θsin⁡θcos⁡θ=1+sec⁡θcosec⁡θ=\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}=1+\sec\theta\cosec\theta

(iv) Left-hand side:

1+sec⁡Asec⁡A=1+cos⁡A\frac{1+\sec A}{\sec A}=1+\cos A

Right-hand side:

sin⁡2A1−cos⁡A=(1−cos⁡A)(1+cos⁡A)1−cos⁡A=1+cos⁡A\frac{\sin^2A}{1-\cos A}=\frac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A

(v) Use 1 − cos A = sin²A/(1 + cos A).

cos⁡A+sin⁡A−1=sin⁡A−(1−cos⁡A)\cos A+\sin A-1=\sin A-(1-\cos A)
=sin⁡A−sin⁡2A1+cos⁡A=sin⁡A(1+cos⁡A−sin⁡A)1+cos⁡A=\sin A-\frac{\sin^2A}{1+\cos A}=\frac{\sin A(1+\cos A-\sin A)}{1+\cos A}
cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=1+cos⁡Asin⁡A=cosec⁡A+cot⁡A\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\frac{1+\cos A}{\sin A}=\cosec A+\cot A

(vi)

1+sin⁡A1−sin⁡A=(1+sin⁡A)2cos⁡2A\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2A}}
=1+sin⁡Acos⁡A=sec⁡A+tan⁡A=\frac{1+\sin A}{\cos A}=\sec A+\tan A

(vii)

sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=sin⁡θ(1−2sin⁡2θ)cos⁡θ(2cos⁡2θ−1)\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}
=sin⁡θ(cos⁡2θ−sin⁡2θ)cos⁡θ(cos⁡2θ−sin⁡2θ)=tan⁡θ=\frac{\sin\theta(\cos^2\theta-\sin^2\theta)}{\cos\theta(\cos^2\theta-\sin^2\theta)}=\tan\theta

(viii)

(sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2(\sin A+\cosec A)^2+(\cos A+\sec A)^2
=sin⁡2A+2+cosec⁡2A+cos⁡2A+2+sec⁡2A=\sin^2A+2+\cosec^2A+\cos^2A+2+\sec^2A
=5+(1+cot⁡2A)+(1+tan⁡2A)=7+tan⁡2A+cot⁡2A=5+(1+\cot^2A)+(1+\tan^2A)=7+\tan^2A+\cot^2A

(ix) Left-hand side:

(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)(\cosec A-\sin A)(\sec A-\cos A)
=1−sin⁡2Asin⁡A⋅1−cos⁡2Acos⁡A=sin⁡Acos⁡A=\frac{1-\sin^2A}{\sin A}\cdot\frac{1-\cos^2A}{\cos A}=\sin A\cos A

Right-hand side:

1tan⁡A+cot⁡A=1sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡Acos⁡A\frac1{\tan A+\cot A}=\frac1{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\sin A\cos A

(x) First expression:

1+tan⁡2A1+cot⁡2A=sec⁡2Acosec⁡2A=tan⁡2A\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A

Second expression:

(1−tan⁡A1−cot⁡A)2=(1−tan⁡A1−1tan⁡A)2\left(\frac{1-\tan A}{1-\cot A}\right)^2=\left(\frac{1-\tan A}{1-\frac1{\tan A}}\right)^2
=(1−tan⁡Atan⁡A−1tan⁡A)2=(−tan⁡A)2=tan⁡2A=\left(\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right)^2=(-\tan A)^2=\tan^2A

Final answer

All ten identities are proved.

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