\[ \large { \left( \tan { x } \right) }^{ 2 }={ \left( \sin { x } \right) }^{ 2 }{ +\left( \sin { x } \right) }^{ 4 }+{ \left( \sin { x } \right) }^{ 6 }+{ \left( \sin { x } \right) }^{ 8 }+\cdots \]

Prove the trigonometric identity above.

\[ \large { \left( \tan { x } \right) }^{ 2 }={ \left( \sin { x } \right) }^{ 2 }{ +\left( \sin { x } \right) }^{ 4 }+{ \left( \sin { x } \right) }^{ 6 }+{ \left( \sin { x } \right) }^{ 8 }+\cdots \]

Prove the trigonometric identity above.

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TopNewest\(\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{S=\dfrac{\sin^2 x}{1-\sin^2 x} = \dfrac{\sin^2x}{\cos^2 x} = \boxed{\tan^2x} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} \) – Nihar Mahajan · 1 year, 1 month ago

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– Rishik Jain · 10 months, 4 weeks ago

The sum of an infinite GP can only be calculated when \(-1<r<1\). However \(\sin^2x\) can be equal to 1. Can you justify?Log in to reply

– Nihar Mahajan · 10 months, 4 weeks ago

It should be specified that \(|\sin x|<1\)Log in to reply

– Joel Yip · 10 months, 4 weeks ago

Well it is already in the range!Log in to reply

@Mehul Arora – Aditya Kumar · 10 months, 4 weeks ago

100% vella! Just likeLog in to reply

It is related to Andrew's recent problem :3 – Nihar Mahajan · 10 months, 4 weeks ago

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– Svatejas Shivakumar · 1 year ago

wow it looks like a pyramid. How long did you take to type and count so many boxes :)?Log in to reply

– Nihar Mahajan · 1 year ago

About 5 mins...Log in to reply

Interesting! – Arnob Roy · 1 year ago

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