Question

if sin theta is equal to a square minus b square divided by a square + b square find 1 + 10 theta cos theta​

Answer 1

Correct Question:

• If sin theta is equal to a square minus b square divided by a square + b square find 1 + tan theta cos theta.

Answer:

• $\large\bold\red{\frac{2 {a}^{2} }{ {a}^{2} + {b}^{2} } }$

Step-by-step explanation:

Given,

• $\sin( \theta) = \frac{{a}^{2} - {b}^{2}}{ {a}^{2} + {b}^{2} }$

To find :

• $\cos( \theta )$

We know that,

$\cos( \theta ) = \sqrt{1 - { \sin }^{2} \theta } \:$

Therefore,

Substituting the values,

We get,

$= > \cos( \theta ) = \sqrt{1 - {( \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } )}^{2} } \\ \\ = > \cos( \theta ) = \sqrt{ \frac{ {( {a}^{2} + {b}^{2}) }^{2} - {( {a}^{2} - {b}^{2} ) }^{2} }{ {( {a}^{2} + {b}^{2} )}^{2} } }$

But,

We know that,

${x}^{2} - {y}^{2} = (x + y)(x - y)$

Therefore,

We get,

$= > \cos( \theta ) = \frac{ \sqrt{( {a}^{2} + {b}^{2} + {a}^{2} - {b}^{2} )( {a}^{2} + {b}^{2} - {a}^{2} + {b}^{2} ) } }{ {a}^{2} + {b}^{2} } \\ \\ = > \cos( \theta ) = \frac{ \sqrt{2 {a}^{2} \times 2 {b}^{2} } }{ {a}^{2} + {b}^{2} } \\ \\ = > \cos( \theta ) = \sqrt{4 {a}^{2} {b}^{2} } \times \frac{1}{ {a}^{2} + {b}^{2} } \\ \\ = > \cos( \theta ) = \frac{2ab}{ {a}^{2} + {b}^{2} }$

Therefore,

We get,

$= > \tan( \theta ) = \frac{ \sin( \theta ) }{ \cos( \theta ) } \\ \\ = > \tan( \theta ) = \frac{ \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } }{ \frac{2ab}{ {a}^{2} + {b}^{2} } } \\ \\ = > \tan( \theta ) = \frac{ {a}^{2} - {b}^{2} }{2ab}$

Thus,

We get,

$1 + \tan( \theta ) \cos( \theta ) = 1 + ( \frac{ {a}^{2} - {b}^{2} }{2ab} )( \frac{2ab}{ {a}^{2} + {b}^{2} } ) \\ \\ = 1 + \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \\ \\ = \frac{ {a}^{2} + {b}^{2} + {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \\ \\ = \large \bold{\frac{2 {a}^{2} }{ {a}^{2} + {b}^{2} } }$