Question

Please solve the question

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Answer 1

Answer:

I tried but can't get.......amigo

Answer 2

Step-by-step explanation:

Questions

if sin theta = 3/4 ; Prove that

$\sqrt{ \frac{ { { \csc(\theta) }^{2} - { \cot( \theta) }^{2} }}{ { \sec(\theta) }^{2} - 1} } = \frac{ \sqrt{7} }{3}$

Formula

${ \csc(\theta) }^{2} - { \cot(\theta) }^{2} = 1$

${ \sec(\theta) }^{2} - 1 = { \tan(\theta) }^{2}$

Solutions

Given

$\sin(\theta ) = \frac{3}{4}$

on squaring both side we get

Sin^2(\theta) = 9/16

Cos^2(\theta) = 1 - 9/16= 7/16

Tan^2(\theta) =

$\frac{ \frac{9}{16} }{ \frac{7}{16} } = \frac{9}{7}$

cot^2(\theta)= 7/9

LHS

on putting the value of above formula

we get

$\sqrt{ \frac{1}{ { \tan(\theta) }^{2} } } \\ \sqrt{ { \cot(\theta) }^{2} } \\ \sqrt{ \frac{7}{9} } \\ \frac{7}{3}$

Hence LHS = RHS

proved