Question

The value of the expression $\tan ({53}^{\circ }+\theta )-\cot ({37}^{\circ }-\theta )+\sin ({48}^{\circ }-\theta )-\cos ({42}^{\circ }+\theta )$ is

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is A 0

$\tan ({53}^{\circ }+\theta )-\cot ({37}^{\circ }-\theta )+\sin ({48}^{\circ }-\theta )-\cos ({42}^{\circ }+\theta )$

$=\tan ({53}^{\circ }+\theta )-\cot [{90}^{\circ }-({53}^{\circ }+\theta \left)\right]+\sin [{90}^{\circ }-({42}^{\circ }+\theta \left)\right]-\cos ({42}^{\circ }+\theta )$

$=\tan ({53}^{\circ }+\theta )-\tan ({53}^{\circ }+\theta )+\cos ({42}^{\circ }+\theta )-\cos ({42}^{\circ }+\theta )(\because \cot ({90}^{\circ }-\theta )=\tan \theta ,\sin ({90}^{\circ }-\theta )=\cos \theta )$

= 0 - 0 = 0

Hence, the correct answer is option (a).