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Which of the following trigonometric identities are true?

sin⁡2(41∘)+sin⁡2(49∘)=1sin⁡2(60∘)−2tan⁡(45∘)−cos⁡2(30∘)=−1sin⁡2(θ)+11+tan⁡2(θ)=1\begin{aligned} & \sin ^2\left(41^{\circ}\right)+\sin ^2\left(49^{\circ}\right)=1 \\ & \sin ^2\left(60^{\circ}\right)-2 \tan \left(45^{\circ}\right)-\cos ^2\left(30^{\circ}\right)=-1 \\ & \sin ^2(\theta)+\frac{1}{1+\tan ^2(\theta)}=1 \end{aligned}

Solution

✅ Correct Option: 2

For statement A:

sin⁡2(41°)+sin⁡2(49°)=sin⁡2(41°)+sin⁡2(90°−41°)\sin^2(41°) + \sin^2(49°) = \sin^2(41°) + \sin^2(90° - 41°)

=sin⁡2(41°)+cos⁡2(41°)=1= \sin^2(41°) + \cos^2(41°) = 1 ✓ [We know that sin⁡2(x)+cos⁡2(x)=1\sin^2(x)+\cos^2(x)=1]

For statement B:

sin⁡2(60°)−2tan⁡(45°)−cos⁡2(30°)\sin^2(60°) - 2\tan(45°) - \cos^2(30°)

=34−2(1)−34=−2≠−1= \frac{3}{4} - 2(1) - \frac{3}{4} = -2 ≠ -1 ✗

For statement C:

sin⁡2(θ)+11+tan⁡2(θ)\sin^2(\theta) + \frac{1}{1 + \tan^2(\theta)}

=sin⁡2(θ)+cos⁡2(θ)= \sin^2(\theta) + \cos^2(\theta) (since 11+tan⁡2(θ)=cos⁡2(θ)\frac{1}{1 + \tan^2(\theta)} = \cos^2(\theta))

=1= 1 ✓

Hence, A and C are true.

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