Question

Question 3
The value of the expression $cosec({75}^{\circ }+\theta )-sec({15}^{\circ }-\theta )-tan({55}^{\circ }+\theta )+cot({35}^{\circ }-\theta )$ is

(A) -1
(B) 0
(C) 1
(D) $\frac{3}{2}$

Answer 1

The answer is (B).
Thinking Process
We see that the given trigonometric angle of the ratio are the reciprocal in the sense of sign. Then, use the following formulae

i) $cosec({90}^{\circ }-\theta )=sec\theta$

ii) $cot({90}^{\circ }-\theta )=tan\theta$

Given, expression $=cosec({75}^{\circ }+\theta )-sec({15}^{\circ }-\theta )-tan({55}^{\circ }+\theta )+cot({35}^{\circ }-\theta )$

$=cosec\lfloor {90}^{\circ }-({15}^{\circ }-\theta )\rfloor -sec({15}^{\circ }-theta)-tan({55}^{\circ }+\theta )+cot\{{90}^{\circ }-({55}^{\circ }+\theta \left)\right\}$

$=sec({15}^{\circ }-\theta )-sec({15}^{\circ }-\theta )-tan({55}^{\circ }+\theta )+tan({55}^{\circ }+\theta )\lfloor \because cosec({90}^{\circ }-\theta )=sec\theta andcot({90}^{\circ }-\theta )\rfloor$

= 0

Hence, the required value of the given expression is 0.