Question

Please solve this question with proper reasoning. Don't spam please.​

Answer 1

Answer:

(A) 3/4

Hope it will help you

Answer 2

Step-by-step explanation:

$\tan( \alpha ) = \frac{1}{ \sqrt{7} } = \frac{p}{b} \\ \\ p = 1 \: \: b = \sqrt{7} \: h = \sqrt{ {1}^{2} + (\sqrt{7} ) {}^{2} } = \sqrt{8 } \\ p = 1 \: b = \sqrt{7} \: h = \sqrt{8} \\ \csc ( \alpha ) = \frac{h}{p} = \frac{ \sqrt{8} }{1} = \sqrt{8} \\ cot( \alpha ) = \frac{b}{p} = \frac{ \sqrt{7} }{1} = \sqrt{7} \\ \frac{ \csc {}^{2} ( \alpha ) - \cot {}^{2} ( \alpha ) }{\csc {}^{2} ( \alpha ) + \cot {}^{2} ( \alpha ) } \\ \frac{( \sqrt{8} ) {}^{2} - ( \sqrt{7}) {}^{2} }{ (\sqrt{8}) {}^{2} + (\sqrt{7}) {}^{2} } \\ \frac{8 - 7}{8 + 7} \\ \frac{1}{15} \: \: \: \: answer$

1. alpha = theta consider karna

2.h=hypotenuse

3.b=base

4.p=perpendicular

5.√p^2+b^2=h (Pythagoras theorem)

check it and ask me more tough question