Introduction to Trigonometry | Exercise 8.2

Question 2

Choose the correct option and justify your choice :

(i) 2tan⁡30∘1+tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = (A) sin⁡60∘\sin 60^\circ (B) cos⁡60∘\cos 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

(ii) 1−tan⁡245∘1+tan⁡245∘=\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = (A) tan⁡90∘\tan 90^\circ (B) 11 (C) sin⁡45∘\sin 45^\circ (D) 00

(iii) sin⁡2A=2sin⁡A\sin 2A = 2 \sin A is true when A=A = (A) 0∘0^\circ (B) 30∘30^\circ (C) 45∘45^\circ (D) 60∘60^\circ

(iv) 2tan⁡30∘1−tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = (A) cos⁡60∘\cos 60^\circ (B) sin⁡60∘\sin 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • To solve these multiple-choice problems, we evaluate each expression by substituting standard trigonometric values (such as tan⁡30∘=13\tan 30^\circ = \dfrac{1}{\sqrt{3}} and tan⁡45∘=1\tan 45^\circ = 1) or by using standard double-angle formulas:
    • sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}
    • cos⁡2A=1−tan⁡2A1+tan⁡2A\cos 2A = \dfrac{1 - \tan^2 A}{1 + \tan^2 A}
    • tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}
  • For equations involving an unknown angle AA, we test each given option by substituting it into both the LHS\text{LHS} and RHS\text{RHS}.

(i) 2tan⁡30∘1+tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = (A) sin⁡60∘\sin 60^\circ (B) cos⁡60∘\cos 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

Step 1 · Evaluate the Expression

Using the double-angle identity sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A = \dfrac{2 \tan A}{1 + \tan^2 A} with A=30∘A = 30^\circ

2tan⁡30∘1+tan⁡230∘=sin⁡(2×30∘)=sin⁡60∘\begin{aligned} \dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} &= \sin (2 \times 30^\circ) \\[0.6em] &= \sin 60^\circ \end{aligned}
Answer

(i) (A) sin⁡60∘\sin 60^\circ

(ii) 1−tan⁡245∘1+tan⁡245∘=\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = (A) tan⁡90∘\tan 90^\circ (B) 11 (C) sin⁡45∘\sin 45^\circ (D) 00

Step 1 · Evaluate the Expression

Diagram 4

Using the identity cos⁡2A=1−tan⁡2A1+tan⁡2A\cos 2A = \dfrac{1 - \tan^2 A}{1 + \tan^2 A} with A=45∘A = 45^\circ

1−tan⁡245∘1+tan⁡245∘=cos⁡(2×45∘)=cos⁡90∘=0\begin{aligned} \dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} &= \cos (2 \times 45^\circ) \\[0.6em] &= \cos 90^\circ \\[0.6em] &= 0 \end{aligned}
Answer

(ii) (D) 00

(iii) sin⁡2A=2sin⁡A\sin 2A = 2 \sin A is true when A=A = (A) 0∘0^\circ (B) 30∘30^\circ (C) 45∘45^\circ (D) 60∘60^\circ

Step 1 · Verify Option (A) A=0∘A = 0^\circ

Substitute A=0∘A = 0^\circ into LHS\text{LHS} and RHS\text{RHS}

LHS=sin⁡(2×0∘)=sin⁡0∘=0\begin{aligned} \text{LHS} &= \sin (2 \times 0^\circ) \\ &= \sin 0^\circ \\ &= 0 \end{aligned} RHS=2sin⁡0∘=2×0=0\begin{aligned} \text{RHS} &= 2 \sin 0^\circ \\ &= 2 \times 0 \\ &= 0 \end{aligned}

Since LHS=RHS=0\text{LHS} = \text{RHS} = 0, the equality holds true when A=0∘A = 0^\circ.

Answer

(iii) (A) 0∘0^\circ

(iv) 2tan⁡30∘1−tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = (A) cos⁡60∘\cos 60^\circ (B) sin⁡60∘\sin 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

Step 1 · Evaluate the Expression

Using the double-angle identity tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A} with A=30∘A = 30^\circ

2tan⁡30∘1−tan⁡230∘=tan⁡(2×30∘)=tan⁡60∘\begin{aligned} \dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} &= \tan (2 \times 30^\circ) \\[0.6em] &= \tan 60^\circ \end{aligned}
Answer

(iv) (C) tan⁡60∘\tan 60^\circ

Common Mistakes
  • Denominator Sign Confusion: In part (i) the denominator has a plus sign (1+tan⁡230∘1 + \tan^2 30^\circ), giving sin⁡60∘\sin 60^\circ, whereas in part (iv) the denominator has a minus sign (1−tan⁡230∘1 - \tan^2 30^\circ), giving tan⁡60∘\tan 60^\circ.
  • Linear Angle Fallacy: Assuming sin⁡2A=2sin⁡A\sin 2A = 2 \sin A is an identity that holds for all angles AA. In reality, it only holds for specific values such as A=0∘A = 0^\circ.

More questions in Exercise 8.2

Q1

Evaluate the following :

(i) sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

(ii) 2tan⁡245∘+cos⁡230∘−sin⁡260∘2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

(iii) cos⁡45∘sec⁡30∘+csc⁡30∘\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}

(iv) sin⁡30∘+tan⁡45∘−csc⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

(v) 5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

Q2

Choose the correct option and justify your choice :

(i) 2tan⁡30∘1+tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = (A) sin⁡60∘\sin 60^\circ (B) cos⁡60∘\cos 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

(ii) 1−tan⁡245∘1+tan⁡245∘=\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = (A) tan⁡90∘\tan 90^\circ (B) 11 (C) sin⁡45∘\sin 45^\circ (D) 00

(iii) sin⁡2A=2sin⁡A\sin 2A = 2 \sin A is true when A=A = (A) 0∘0^\circ (B) 30∘30^\circ (C) 45∘45^\circ (D) 60∘60^\circ

(iv) 2tan⁡30∘1−tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = (A) cos⁡60∘\cos 60^\circ (B) sin⁡60∘\sin 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

Q3

If tan⁡(A+B)=3\tan (A + B) = \sqrt{3} and tan⁡(A−B)=13\tan (A - B) = \dfrac{1}{\sqrt{3}}; 0∘<A+B≤90∘0^\circ < A + B \le 90^\circ; A>BA > B, find AA and BB.

Q4

State whether the following are true or false. Justify your answer.

(i) sin⁡(A+B)=sin⁡A+sin⁡B\sin (A + B) = \sin A + \sin B.

(ii) The value of sin⁡θ\sin \theta increases as θ\theta increases.

(iii) The value of cos⁡θ\cos \theta increases as θ\theta increases.

(iv) sin⁡θ=cos⁡θ\sin \theta = \cos \theta for all values of θ\theta.

(v) cot⁡A\cot A is not defined for A=0∘A = 0^\circ.

← Back to Introduction to Trigonometry