Question

The value of$\frac{tan55^{0}}{cot35^{0} }+cot 1^{0}cot2^{0}+cot3^{0}....cot90^{0}$, is
(a)−2
(b)2
(c)1
(d)0

Answer 1

SOLUTION :  

The correct option is  (c)  : 1

Given : tan 55° / cot 35° +  cot 1° cot 2° cot 3° ....... cot 90°.

= tan (90° - 35°)/cot 35° + cot (90° - 89°) cot (90° - 88°) cot (90° - 87°) ………. cot 87° cot 89° cot 90°

= cot 35° / cot 35° + tan 89° tan 88°tan 87°…...cot 87° cot 89° cot 90°

[ tan (90° - θ) = cot θ]

= 1 +( tan 89° cot 89°)( tan 88° cot 88° )….. cot 90°  

= 1 + 1 × 1 …… × 0

[ cot θ × tan θ = 1, cot 90° = 0]

= 1 + 0  

= 1  

Hence, the value of tan 55° / cot 55° +  cot 1° cot 2° cot 3° ....... cot 90° is 1.

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Hey there!

In this type of questions,the answer will always come 1,so there is no particular method to do it.