Question

If sinA , cos A and tanA are in geometric progression then cot ^6 A - cot ^2 A is ___________
a) 2 b) 3 c)4 d) 1
explain

Answer 1

Hey !

Here is your answer !!

➡Given !!

➡sinA, cosA and tanA are in GP, then

Then b2 = ac

➡CosA/SinA = tanA/cosA

➡Cos^2A = tanA.sinA (Cross-multiply )

➡Cos^3A = sin^2A (Simplifying using tanA = sinA/cosA)......Equation 1

Now !!

➡CosA/SinA = tanA/cosA (Given)

➡CotA = 1/cotA .1/cosA

➡Cot^2A = 1/cosA = secA (Cross-

multiplying).............Equation 2

Now !!
➡cot^6A = sec^3A (On cubing both sides of Equation 2 ) .... Equation 3

Now,=>>cot^6A - cot^2A = sec^3A - secA [ Subtract equation 3 from equation 2 ]

➡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..............Equation 4

Substituting equation 1 in equation 4

______________
We know, sin^2A = cos^3A

So,
Cot^6A - Cot^2A = 1

Option d

Hope it helps !!

Answer 2

Answer:

Step-by-step explanation:

Cos A/sin A= tan A / cos A

Cos ^2 A =tan A sin A

= sin^2 A /cos ^2 A

Cos ^3 A = sin ^2 A

From question

Cot^6 A -cot ^2 A = cos ^6 A / sin ^6A

-cos ^2 A/ sin^2A

= (1-cos^2A)/sin^2A

= 1