Question

If sin A , cos A , tan A are in Geometric Progression . then \[\cot^6 a-\cot^2a\]= ??

Answer 1

\[\cos^2A=\sin A \tan A=\frac{\sin ^2A}{\cos A}\]
\[\cot^2A=\frac{1}{\cos A}\]
\[\cot^6A-\cot^2A=\frac{1}{\cos^3A} -\frac{1}{ \cos A}\]

Answer 2

simplify it to get an integer value!

Answer 3

how did u get \[\huge \cot^6a=\frac{ 1 }{ \cos^2a }\]?

Answer 4

\[\frac{\cos A-\cos ^3A}{\cos^4A}=\frac{\cos A(1-\cos^2A)}{\cos^4A}=\frac{ (\sin^2A)}{\cos^3A}\]

Answer 5

please answer my query !

Answer 6

since
\[\cot ^2A=\frac{1}{\cos A}\]
cube everyside...
\[(\cot ^2A)^3=(\frac{1}{\cos A})^3\implies \cot ^6A=\frac{1}{\cos^3 A}\]\]

Answer 7

simplify 1/cos^3 A - 1/cosA and use cos^2 A = sinA tanA to get your finals answer

Answer 8

i cant find a number 4 that identity...

Answer 9

There exists an integer for that.
Because we know the value of sin^2 A / cos^3 A

Answer 10

im getting 1

Answer 11

1/cos^3 A - 1/cosA

1 - cos^2A
------------
cos^3A

Tan^2A
--------
cos A

cos A
---- (cos^2A = sinA tanA => tan^2a = cosA)
cos A

1