Question

See the image and plss answer the question ​

Answer 1

Answer:-

(Theta is taken as "A".)

Given:

$\sf{\frac{ \sin\:A- \cos\:A}{ \sin\:A+ \cos\:A } + \frac{ \sin\:A + \cos\:A }{ \sin\:A - \cos\:A} = \frac{2}{2 { \sin }^{2} A - 1}}$

Taking LCM in LHS,

$\sf \implies\frac{ {( \sin\:A - \cos\:A) }^{2} + {( \sin\:A + \cos\:A) }^{2} }{( \sin\:A + \cos\:A)( \sin\:A - \cos\:A ) }$

[ (a + b)(a + b) = (a + b)² and (a - b)(a - b) = (a - b)² ]

$\sf \implies \: \frac{ { \sin }^{2} A + { \cos}^{2}A - 2 \sin\:A \cos\:A + { \sin}^{2} A + { \cos }^{2} A + 2 \sin\:A\cos\:A }{ { \sin }^{2} A - { \cos }^{2} A}$

[ (a + b)² = a² + b² + 2ab ;

(a - b)² = a² + b² - 2ab,

(a + b)(a - b) = a² - b²]

$\sf \implies \: \frac{1 + 1}{ { \sin }^{2}A - (1 - { \sin}^{2}A) }$

[ Sin² ∅ + Cos² ∅ = 1 → Cos² ∅ = 1 -

Sin²∅ ]

$\sf \implies \: \frac{2}{ {{ \sin }^{2} A} + { \sin }^{2} A - 1 } \\ \\ \sf \implies \: \frac{2}{2 { \sin }^{2}A - 1 } = RHS.$

Hence, Proved.

Answer 2

Answer:

refer the attachment for the solution.

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