Question

answer please please​

Answer 1

Step-by-step explanation:

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Answer 2

Explanation :-

We have -

$\to \sf{\tan \theta + \sin \theta = p \: } \: \\ \bf{ and }\: \\ \to \sf{ \tan \theta - \sin \theta = q}$

taking LHS ( left hand side)

$\implies \sf{ {p}^{2} - {q}^{2} } \\ \\ \implies \sf{ (p - q)(p + q)} \\ \\ \implies \: \sf{(\tan \theta + \sin \theta - \tan \theta + \sin \theta) } \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\times ( \tan \theta + \sin \theta + \tan \theta - \sin \theta) \\ \\ \implies \: 2 \sin \theta \times 2 \tan \theta \\ \\ \implies \:4 \: \tan \theta \: \sin \theta \:$

now taking RHS ( right hand side)

$\to \sf{ 4 \sqrt{pq} } \\ \\ \implies \: 4 \sqrt{( \sf{\tan \theta + \sin \theta )(\sf{\tan \theta - \sin \theta )}} \:} \\ \\ \implies \: 4 \sqrt{ { \tan}^{2} \theta - { \sin}^{2} \theta } \\ \\ \implies \: 4 \sqrt{ \frac{ { \sin}^{2} \theta}{ { \cos}^{2} \theta } - { \sin}^{2} \theta } \\ \\ \implies \: 4 \sin \theta \sqrt{ \frac{1}{ { \cos}^{2} \theta } - 1} \\ \\ \because \boxed{ \frac{1}{ \cos \theta} = \sec \theta } \\ \\ \implies \: 4 \sin \theta \sqrt{ { \sec}^{2} \theta - 1 } \\ \\ \because \boxed{ { \sec}^{2} \theta - 1 = { \tan}^{2} \theta \: } \\ \\ \implies \: 4 \sin \theta \sqrt{ { \tan}^{2} \theta} \\ \\ \implies \: 4 \tan \theta \: \sin \theta$

LHS = RHS

Hence proved .