Question

pls solve the ques 37 plsssss. i want sol .​

Answer 1

Answer:

Pls go through the attachment

Identity used =sin (90-a)=cosa

Sorry cosec 45=root2

Bblegend

Answer 2

$\huge\bigstar\huge\mathcal{\underline{ \underline{QUESTION}}}\red\bigstar$

$sin3 \theta = cos( 2\theta + 15 \degree)where \: 3 \theta \: \\ and( \theta - 6 \degree) \: are \: both \: acute \: angles \: \\ find \: the \: value \: of \: tan \theta \: and \: cosec3 \theta.$

$\huge\red\bigstar\huge\mathcal{\underline{ \underline{SOLUTION}}}\bigstar$

$\boxed{}$

$\implies sin3 \theta = cos( 2\theta + 15 \degree) \\ \implies sin3 \theta = sin(90 \degree - (2 \theta + 15 \degree)) \\ \implies 3 \theta = 90 \degree - 2 \theta - 15 \degree \\ \implies 3 \theta + 2 \theta \: = 75 \degree \\ \implies 5 \theta \: = 75 \degree \\ \implies \theta = \frac{\cancel{75 \degree}}{\cancel{5}} \\ \implies \theta = 15 \degree$

$now ...to \: find \:tan2 \theta \: and \: cosec3 \theta$

now....

$\rightarrow tan 2\theta = tan \: 2(15 \degree) \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = tan30 \degree \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{ \sqrt{3} }$

and

$\rightarrow cosec3 \theta = cosec \: 3(15 \degree) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = cosec \: 45 \degree \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{2}$

$\large\mathfrak{...hope\: this \:helps\: you.....}$