`cot16^0cot44^0+cot44^0cot76^0-cot76^0cot16^0=` 1 (b) 2 (c) 3 (d) 4

cot16^0cot44^0+cot44^0cot76^0-cot76^0cot16^0 = (a) 1 (b) 2 (c) 3 (d) 4

cot 6^0 cot 42^0 cot 66^0 cot 78^0=1

The value of (tan55^0)/(cot35^0)+cot1^0cot2^0cot3^0cot90^0,i s -2 (b) 2 (c) 1 (d) 0

Value of "cot"40^0dotcot 20^0(4sin 10^0-1) is -1 (b)1 (c) 2 (d) 0

Without using trigonometric tables, evaluate the following: (sec37^0)/(cos e c53^0)+2cot15^0cot25^0cot45^0cot75^0cot65^0(sin^2 18^0+sin^2 72^0)

The value of (3+cot76^@cot16^@)/(cot76^@+cot16^@) is

Value of (3+cot80^0cot20^0)/(cot80^0+cot20^0) is equal to (a) cot20^0 (b) tan50^0 (c) cot50^0 (d) cotsqrt(20^0)

Prove that: cotA+"cot"(60^0+A)-cot(60^0-A)=3cot3A

Prove that: cotA+"cot"(60^0+A)-cot(60^0-A)=3cot3A

If A =tan6^0tan42^0 and B=cot66^0cot78^0 , then