Question

if sinA,cosA,tanA are in GP then find cot power 6 - cot²=?​

Answer 1

If sinA, cosA and tanA are in GP, then 

CosA/SinA = tanA/cosA 

Cos^2A = tanA.sinA (Cross-multiplying) 

Cos^3A = sin^2A

(Simplifying using tanA = sinA/cosA)------------(1)

CosA/SinA = tanA/cosA (Given) 

CotA = 1/cotA .1/cosA 

Cot^2A = 1/cosA = secA (Cross-multiplying)-------(2) 

Then, cot^6A = sec^3A (On cubing both sides of 2)----------(3) 

Now, equation(3) -equation(2)

= cot^6A - cot^2A

= sec^3A - secA 

=secA(sec^2A - 1) [taking secA common term outside] 

=secA.tan^2A [from

Identity sec^2A - tan^2A = 1] 

=1/cosA . Sin^2A/cos^2A 

=sin^2A/cos^3A..............4 

All the best

Substituting 1 in 4 

We know, sin^2A = cos^3A 

So, 

Cot^6A - Cot^2A = 1