Basic trigo - III

Geometry Level 3

If sin ⁡ A , cos ⁡ A \sin { A } ,\cos { A } and tan ⁡ A \tan { A } are in geometric progression, then cot ⁡ 6 A − cot ⁡ 2 A \cot ^{ 6 }{ A } -\cot ^{ 2 }{ A } is equal to??

This question is a part of the set: Basic Trigo

1 0 -1 csc ⁡ 2 A \csc ^{ 2 }{ A }

Krishna Ramesh by Krishna Ramesh , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Krishna Ramesh
May 21, 2014

Here's my method(may be slightly longer than Vinay's):

Since they are in G.P, we have cos ⁡ 2 A = sin ⁡ A tan ⁡ A \cos ^{ 2 }{ A } =\sin { A } \tan { A }

now, cot ⁡ 2 A = cos ⁡ 2 A sin ⁡ 2 A = sin ⁡ A tan ⁡ A sin ⁡ 2 A = sec ⁡ A \cot ^{ 2 }{ A } =\frac { \cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } =\frac { \sin { A } \tan { A } }{ \sin ^{ 2 }{ A } } =\sec { A }

⇒ cot ⁡ 4 A = sec ⁡ 2 A ⇒ cot ⁡ 4 A = 1 + tan ⁡ 2 A \Rightarrow \cot ^{ 4 }{ A } =\sec ^{ 2 }{ A } \\ \Rightarrow \cot ^{ 4 }{ A } =1+\tan ^{ 2 }{ A }

so, cot ⁡ 6 A − cot ⁡ 2 A = cot ⁡ 2 A ( cot ⁡ 4 A − 1 ) = cot ⁡ 2 A ( 1 + tan ⁡ 2 A − 1 ) = cot ⁡ 2 A ( tan ⁡ 2 A ) = 1 \cot ^{ 6 }{ A } -\cot ^{ 2 }{ A } =\cot ^{ 2 }{ A\left( \cot ^{ 4 }{ A } -1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cot ^{ 2 }{ A } (1+\tan ^{ 2 }{ A } -1)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cot ^{ 2 }{ A\left( \tan ^{ 2 }{ A } \right) } =1

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Vinay Sipani
May 20, 2014

They are in GP = > C o s 2 A = S i n A . T a n A => Cos^2A = SinA.TanA = > C o t 2 A = s e c A => Cot^2A=secA = > C o s 3 A = s i n 2 A => Cos^3A=sin^2A = > C o t 6 A − C o t 2 A = s e c 3 A − C o t 2 A => Cot^6A-Cot^2A= sec^3A-Cot^2A = c o s e c 2 A − c o t 2 A = cosec^2A - cot^2A = 1 =1

1 Helpful 0 Interesting 0 Brilliant 0 Confused

same method and got it right.

Sai Ram - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...