Question

solve the question with explanation !!!! ​

Answer 1

✮ Question ✮

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Prove : $\tt {\sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } = sec \: \theta + tan \: \theta}$

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Answer :-

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✮ Solving LHS ✮

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$\sf { \sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } }$

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✮ Rationalizing ✮

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$\leadsto \sf { = \sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} \times \dfrac{1 + sin \: \theta}{1 + sin \: \theta} }}$

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$\leadsto \sf {= \sqrt{ \dfrac{ {( 1 + sin \: \theta)}^{2} }{1 - {sin}^{2}\theta } }}$

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Identity : sin²∅ + cos²∅ = 1

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$\leadsto { \sf {=\sqrt{ \dfrac{{(1 + sin \: \theta)}^{2} }{{cos}^{2} \theta} }} }$

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$\leadsto \sf {= \sqrt{{ \bigg( \dfrac{1 + sin \: \theta}{cos \: \theta} \bigg)}^{2} }}$

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$\leadsto \sf{ =\dfrac{1}{cos \: \theta} + \dfrac{sin \: \theta}{cos \: \theta} }$

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$\leadsto \sf {=sec \: \theta + tan \: \theta }$

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Hence, Proved !

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Answer 2

We need to prove ,

• $\displaystyle\sf\sqrt{\dfrac{1+sin\theta}{1-sin\theta}} = sec\theta-tan\theta$

Taking LHS and proving LHS = RHS

Rationalizing the denominator . To rationalize we need to multiply and divide with $\sf (1+sin\theta)$

$\displaystyle \longrightarrow\sf \sqrt{\dfrac{(1+sin\theta)}{(1-sin\theta)}\times\dfrac{(1+sin\theta)}{(1+sin\theta)}}$

$\displaystyle\sf\longrightarrow\sqrt{\dfrac{(1+sin\theta)\times(1+sin\theta)}{(1-sin\theta)\times(1+sin\theta)}}$

• $\sf (a+b)(a+b)=(a+b)^2$
• $\sf (a-b)(a+b) = a^2-b^2$

$\displaystyle\sf\longrightarrow\sqrt{\dfrac{(1+sin\theta)^2}{1^2-sin^2\theta}}$

$\displaystyle\sf\longrightarrow\dfrac{\sqrt{(1+sin\theta)^2}}{\sqrt{1-sin^2\theta}}$

$\displaystyle\sf\longrightarrow\dfrac{1+sin\theta}{\sqrt{cos^2\theta}}$

$\displaystyle\sf\longrightarrow\dfrac{1+sin\theta}{cos\theta}$

$\displaystyle\sf\longrightarrow\dfrac{1}{cos\theta}+\dfrac{sin\theta}{cos\theta}$

$\displaystyle\longrightarrow\textbf{\textsf{sec \theta +tan \theta }}$

• LHS = RHS

Hence proved $\green{\boldsymbol{\checkmark}}$