Question

If
$\sec \theta \: + \tan \theta = \bf \: x$
Then prove that
$\sf \: sin \theta = \dfrac{ {x}^{2} - 1 }{ {x}^{2} + 1}$
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Answer 1

See this Attachment.

here is your answer.

Answer 2

GIVEN :-

• Sec∅ + tan∅ = x.

TO PROVE :-

• Sin∅ = (x² - )/(x² + 1).

SOLUTION :-

Taking the RHS part,

$\implies \sf \: \dfrac{x ^{2} - 1 }{x ^{2} + 1 }$

Substitute the value of x,

$\implies \sf \: \dfrac{( \sec \theta + \tan \theta) ^{2} - 1 }{ ( \sec \theta + \tan \theta) ^{2} + 1}$

$\implies \sf \: \dfrac{ \sec ^{2} ( \theta) + \tan ^{2} ( \theta) + 2 \sec( \theta) \tan(theta) - 1 }{\sec ^{2} ( \theta) + \tan ^{2} ( \theta) + 2 \sec( \theta) \tan(theta) + 1 }$

Now as we know that , sec²A - 1 = tan²A.

$\implies \sf \: \dfrac{ \tan ^{2} ( \theta) + \tan ^{2} ( \theta) + 2 \sec( \theta) \tan( \theta) }{ \sec ^{2} ( \theta) + \sec ^{2} ( \theta) + 2 \sec( \theta) \tan( \theta)}$

$\implies \sf \: \dfrac{2 \tan ^{2} ( \theta) + 2 \sec( \theta) \tan( \theta)}{2 \sec ^{2} ( \theta) + 2 \sec( \theta) \tan( \theta)}$

Taking 2tanA as a common in numerator and 2secA in denominator,

$\implies \sf \: \dfrac{2 \tan( \theta) \cancel{\bigg( \tan( \theta) + \sec( \theta) \bigg)} }{2 \sec( \theta) \cancel{ \bigg( \tan( \theta) + \sec( \theta) \bigg)}}$

$\implies \sf \: \dfrac{2 \tan( \theta)} {2 \sec( \theta) }$

$\implies \sf \: \dfrac{ \tan( \theta) }{ \sec( \theta) }$

As we know that , tanA = sinA/cosA and secA = 1/cosA,

$\implies \sf \: \dfrac{ \dfrac{ \sin( \theta) }{ \cos( \theta) } }{ \dfrac{1}{ \cos( \theta) } }$

$\implies \sf \: \dfrac{\sin( \theta) }{ \cos( \theta) } \times \cos( \theta)$

$\implies \sf \: \sin( \theta)$

Hence proved.