Question

Solve this, genius.... ​

Answer 1

Your attachment is being prepared

Here's your answer ....

Hope it helps ...

Please mark brainliest if it helped

Answer 2

Question :

$\tt \: If \tan \theta = \frac{2mn}{ {m}^{2} - {n}^{2} } \\ \tt \: find \: the \: other \: t \: ratio \: of \: \theta.$

Solution :

We have,

The value of tan theta,

$\tt\dagger \: \: \: \tan \theta = \frac{Perpendicular}{Base} = \frac{2mn}{ {m}^{2} - {n}^{2} }$

$\rule{90}1$

Let's Apply Pythagoras theorem,

$\tt\longmapsto{H }^{2} ={ P }^{2} + {B }^{2}$

Here,

• P = 2mn.
• B = m² - n².
• H = ?

$\tt\longmapsto {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\tt \longmapsto {AC}^{2} = {2mn}^{2} + { ({m}^{2} - {n}^{2}) }^{2} .$

$\tt \longmapsto {AC}^{2} = 4 {m}^{2} {n}^{2} + {m}^{4} + {n}^{4} - 2 {m}^{2} {n}^{2}$

$\tt\longmapsto {AC}^{2} = {m}^{2} + {n}^{4} + 2 {m}^{2} {n}^{2}$

$\tt\longmapsto {AC}^{2} = {( {m}^{2} + {n}^{2} )}^{2}$

$\tt\longmapsto {AC} = {m}^{2} + {n}^{2} .$

$\rule{120}1$

Now, We got all sides.

• AB = 2mn.
• BC = m²- n².
• AC = m² + n².

Let try to Find T - Ratio of theta.

$\tt\dashrightarrow \sin \theta = \frac{P}{H} = \frac{2mn}{ {m}^{2} + {n}^{2} }$

$\tt\dashrightarrow \cos \theta = \frac{B}{H} = \frac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }$

$\tt\dashrightarrow \tan \theta = \frac{P}{B} = \frac{2mn}{ {m}^{2} - {n}^{2} }$

$\tt\dashrightarrow \csc \theta = \frac{H}{P} = \frac{ {m}^{2} + {n}^{2} }{2mn}$

$\tt\dashrightarrow \sec \theta = \frac{H}{B} = \frac{ {m}^{2} + {n}^{2} }{ {m}^{2} - {n}^{2} }$

$\tt\dashrightarrow \cot \theta = \frac{B}{P} = \frac{ {m}^{2} - {n}^{2} }{2mn}$

$\rule{200}2$

Note : Diagram provide in attachment.