Skip to main contentSkip to solution

Which of the following trigonometric identities are true?

sin2(41)+sin2(49)=1sin2(60)2tan(45)cos2(30)=1sin2(θ)+11+tan2(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:

sin2(41°)+sin2(49°)=sin2(41°)+sin2(90°41°)\sin^2(41°) + \sin^2(49°) = \sin^2(41°) + \sin^2(90° - 41°)

=sin2(41°)+cos2(41°)=1= \sin^2(41°) + \cos^2(41°) = 1 ✓ [We know that sin2(x)+cos2(x)=1\sin^2(x)+\cos^2(x)=1]

For statement B:

sin2(60°)2tan(45°)cos2(30°)\sin^2(60°) - 2\tan(45°) - \cos^2(30°)

=342(1)34=21= \frac{3}{4} - 2(1) - \frac{3}{4} = -2 ≠ -1

For statement C:

sin2(θ)+11+tan2(θ)\sin^2(\theta) + \frac{1}{1 + \tan^2(\theta)}

=sin2(θ)+cos2(θ)= \sin^2(\theta) + \cos^2(\theta) (since 11+tan2(θ)=cos2(θ)\frac{1}{1 + \tan^2(\theta)} = \cos^2(\theta))

=1= 1

Hence, A and C are true.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question