If sin A- cos A and tan A are in geometric progression- then 6 A- 2 A is equal toA- aB- None of theseC- 1D- -1

The correct option is C 1Since sin⁡ A,cos⁡ A and tan⁡ A are in G.P., we have cos^2⁡A=sin A tan A ⇒ cos^3⁡A=sin^2⁡A ⇒ cot^2⁡A=sec ⁡A ⇒ cot^4⁡A=1+tan^2⁡A ⇒ c ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If sin A, cos...

Question

If $s i n A, c o s A a n d t a n A$ are in geometric progression, then $c o t^{6} A - c o t^{2} A$ is equal to

-1

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None of these

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Solution

The correct option is C 1
Since $s i n A, c o s A a n d t a n A$ are in G.P., we have
$c o s^{2} A = s i n A t a n A \Rightarrow c o s^{3} A = s i n^{2} A \Rightarrow c o t^{2} A = s e c A$
$\Rightarrow c o t^{4} A = 1 + t a n^{2} A \Rightarrow c o t^{6} A - c o t^{2} A = 1$

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If sin A,cos A and tan A are in geometric progression, then cot6A−cot2A is equal to

The correct option is C 1Since sin A,cos A and tan A are in G.P., we have cos2A=sin A tan A⇒cos3A=sin2A⇒cot2A=sec A ⇒cot4A=1+tan2A⇒cot6A−cot2A=1

If $s i n A, c o s A a n d t a n A$ are in geometric progression, then $c o t^{6} A - c o t^{2} A$ is equal to

The correct option is C 1
Since $s i n A, c o s A a n d t a n A$ are in G.P., we have
$c o s^{2} A = s i n A t a n A \Rightarrow c o s^{3} A = s i n^{2} A \Rightarrow c o t^{2} A = s e c A$
$\Rightarrow c o t^{4} A = 1 + t a n^{2} A \Rightarrow c o t^{6} A - c o t^{2} A = 1$