Introduction to Trigonometry | Exercise 8.3

Question 1

Express the trigonometric ratios sin⁡A\sin A, sec⁡A\sec A and tan⁡A\tan A in terms of cot⁡A\cot A.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • To express trigonometric ratios in terms of cot⁡A\cot A, we use the fundamental Pythagorean identities and reciprocal relations:
    • Reciprocal identity: tan⁡A=1cot⁡A\tan A = \dfrac{1}{\cot A} and sin⁡A=1csc⁡A\sin A = \dfrac{1}{\csc A}
    • Pythagorean identities: csc⁡2A=1+cot⁡2A\csc^2 A = 1 + \cot^2 A and sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A
  • For acute angle AA, all square roots are taken with positive values.

Step 1 · Express sin⁡A\sin A in terms of cot⁡A\cot A

Diagram 1

Using the identity csc⁡2A−cot⁡2A=1\csc^2 A - \cot^2 A = 1

csc⁡2A=1+cot⁡2A\csc^2 A = 1 + \cot^2 A

Taking square root on both sides

csc⁡A=1+cot⁡2A\csc A = \sqrt{1 + \cot^2 A}

Since sin⁡A=1csc⁡A\sin A = \dfrac{1}{\csc A}

sin⁡A=11+cot⁡2A\sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}}

Step 2 · Express sec⁡A\sec A in terms of cot⁡A\cot A

Using the identity sec⁡2A−tan⁡2A=1\sec^2 A - \tan^2 A = 1

sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A

Substitute tan⁡A=1cot⁡A\tan A = \dfrac{1}{\cot A}

sec⁡2A=1+(1cot⁡A)2=1+1cot⁡2A=cot⁡2A+1cot⁡2A\begin{aligned} \sec^2 A &= 1 + \left(\dfrac{1}{\cot A}\right)^2 \\[0.6em] &= 1 + \dfrac{1}{\cot^2 A} \\[0.6em] &= \dfrac{\cot^2 A + 1}{\cot^2 A} \end{aligned}

Taking square root on both sides

sec⁡A=1+cot⁡2Acot⁡2A=1+cot⁡2Acot⁡2A=1+cot⁡2Acot⁡A\begin{aligned} \sec A &= \sqrt{\dfrac{1 + \cot^2 A}{\cot^2 A}} \\[0.8em] &= \dfrac{\sqrt{1 + \cot^2 A}}{\sqrt{\cot^2 A}} \\[0.8em] &= \dfrac{\sqrt{1 + \cot^2 A}}{\cot A} \end{aligned}

Step 3 · Express tan⁡A\tan A in terms of cot⁡A\cot A

Using the reciprocal relation

tan⁡A=1cot⁡A\tan A = \dfrac{1}{\cot A}
Answer

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

Common Mistakes
  • Denominator Square Root: Forgetting to simplify cot⁡2A\sqrt{\cot^2 A} to cot⁡A\cot A in the denominator of sec⁡A\sec A.
  • Sign Ambiguity: Writing ±\pm before square roots; for acute angle AA (Class 10), all trigonometric ratios are strictly positive.
  • Identity Mix-Up: Incorrectly writing sec⁡2A=1−tan⁡2A\sec^2 A = 1 - \tan^2 A instead of sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A.

More questions in Exercise 8.3

Q1

Express the trigonometric ratios sin⁡A\sin A, sec⁡A\sec A and tan⁡A\tan A in terms of cot⁡A\cot A.

Q2

Write all the other trigonometric ratios of ∠A\angle A in terms of sec⁡A\sec A.

Q3

Choose the correct option. Justify your choice.

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

(A) 11 (B) 99 (C) 88 (D) 00

(ii) (1+tan⁡θ+sec⁡θ)(1+cot⁡θ−csc⁡θ)=(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \csc \theta) =

(A) 00 (B) 11 (C) 22 (D) −1-1

(iii) (sec⁡A+tan⁡A)(1−sin⁡A)=(\sec A + \tan A) (1 - \sin A) =

(A) sec⁡A\sec A (B) sin⁡A\sin A (C) csc⁡A\csc A (D) cos⁡A\cos A

(iv) 1+tan⁡2A1+cot⁡2A=\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =

(A) sec⁡2A\sec^2 A (B) −1-1 (C) cot⁡2A\cot^2 A (D) tan⁡2A\tan^2 A

Q4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}

(ii) cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A

(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ cosec θ\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta [Hint : Write the expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta]

(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A} [Hint : Simplify LHS and RHS separately]

(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec A+cot⁡A\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A, using the identity cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A.

(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\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(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A

(ix) (cosec A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A(\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A} [Hint : Simplify LHS and RHS separately]

(x) (1+tan⁡2A1+cot⁡2A)=(1−tan⁡A1−cot⁡A)2=tan⁡2A\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A

← Back to Introduction to Trigonometry