Question

$\tt{\large{\underbrace{ \red{Question:-}}}}$

$\\\sf{Prove \: that:-}$
$\\ \rm{ \rightarrow{ \sqrt{ \dfrac{1 - sin \theta}{1 + sin \theta} } = sec \theta - tan \theta}}$
$\\ \rm{ \rightarrow{ \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } = cot \theta + cosec \theta}}$

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Answer 1

Answer:-

We have to prove:

$1) \: \: \: \: \sqrt{ \dfrac{1 - \sin( \theta) }{1 + \sin( \theta) } } = \sec( \theta) - \tan( \theta)$

Rationalising the denominator in LHS we get ,

$\implies \sf\sqrt{ \dfrac{1 - \sin( \theta) }{1 + \sin( \theta) } \times \dfrac{1 - \sin( \theta) }{1 - \sin (\theta)} } = \sec( \theta) - \tan( \theta) \\ \\ \\ \implies \sf \sqrt{ \frac{ \big(1 - \sin( \theta) \big) ^{2} }{ \big(1 + \sin( \theta) \big) \big(1 - \sin( \theta) \big)} } = \sec( \theta) - \tan( \theta)$

using (a + b)(a - b) = a² - b² in LHS we get;

$\implies \sf \: \sqrt{ \dfrac{ \big(1 - \sin( \theta) \big) ^{2} }{ {1} - { \sin }^{2} ( \theta) } } = \sec( \theta) - \tan( \theta)$

Using the identity cos² θ = 1 - sin² θ we get,

$\implies \sf \: \sqrt{ \frac{ \big(1 - \sin( \theta) \big) ^{2} }{ { \cos}^{2} ( \theta)} } = \sec( \theta) - \tan( \theta) \\ \\ \\ \implies \sf \: \sqrt{ \bigg( \frac{1 - \sin( \theta)}{ \cos( \theta)} \bigg) ^ { 2}} =\sec( \theta) - \tan( \theta) \\ \\ \\ \implies \sf\frac{1 - \sin( \theta)}{ \cos( \theta)} = \sec( \theta) - \tan( \theta) \\ \\ \\ \implies \sf\frac{1 }{\cos( \theta)} - \frac{ \sin( \theta)}{ \cos (\theta)} = \sec( \theta) - \tan( \theta)$

Using 1/cos θ = sec θ and sin θ/cos θ = tan θ we get,

$\implies \sec( \theta) - \tan( \theta) = \sec( \theta) - \tan( \theta)$

Hence, Proved.

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$2) \: \: \: \: \sqrt{ \dfrac{ \sec( \theta) + 1}{ \sec( \theta) - 1} } = \cot( \theta) + \csc( \theta)$

Again rationalising the denominator we get,

$\implies \sf \: \sqrt{ \dfrac{ \sec( \theta) + 1 }{ \sec( \theta) - 1} \times \dfrac{\sec( \theta) + 1}{\sec( \theta) + 1} } = \cot( \theta) + \csc( \theta) \\ \\ \\ \implies \sf \: \sqrt{ \frac{{ \big( \sec( \theta) + 1 \big)}^{2}}{ { \sec }^{2}( \theta) - 1 } } = \cot( \theta) + \csc( \theta)$

using tan² θ = sec² θ - 1 in LHS we get,

$\implies \sf\sqrt{ \frac{{ \big( \sec( \theta) + 1 \big)}^{2}}{ { \tan }^{2}( \theta) } } = \cot( \theta) + \csc( \theta) \\ \\ \\ \implies \sf \sqrt{ \bigg( \frac{ \sec( \theta) + 1 }{ \tan (\theta)} \bigg) ^{2} } = \cot( \theta) + \csc( \theta) \\ \\ \\ \: \implies \sf \frac{ \sec( \theta) + 1 }{ \tan (\theta)} = \cot( \theta) + \csc( \theta) \\ \\ \\ \implies \sf \: \frac{ \sec( \theta) }{ \tan( \theta)} + \frac{1}{ \tan (\theta)} = \cot( \theta) + \csc( \theta)$

Using sec θ = 1/cos θ ; tan θ = sin θ/cos θ and 1/tan θ = cot θ in LHS we get,

$\implies \sf \: \dfrac{ \frac{1}{ \cancel{\cos( \theta) }} }{ \frac{ \sin( \theta) }{ \cancel{\cos (\theta)}} } + \cot( \theta) = \csc( \theta) + \cot( \theta) \\ \\ \\ \implies \sf \frac{1}{ \sin( \theta) } + \cot( \theta) = \csc( \theta) + \cot( \theta)$

Using cosec θ = 1/sin θ we get,

$\implies \csc( \theta) + \cot( \theta) = \csc( \theta) + \cot( \theta)$

Hence, Proved.

Answer 2

$\dag{ \large{ \underline{ \underline{ \sf{ \pmb{Question}}}}}}$

$\\\sf{Prove \: that:-}$

$\\ \rm{ \bold{ \rightarrow{ \sqrt{ \dfrac{1 - sin \theta}{1 + sin \theta} } = sec \theta - tan \theta}}}$

$\\ \rm{ \bold{ \rightarrow{ \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } = cot \theta + cosec \theta}}}$

$\\ \dag{ \large{\underline{ \underline{ \sf{ \pmb{Required \: answers}}}}}}$

$\dag{ \underline{ \underline{ \sf{Question \: (1)}}}}$

$\\ \rm{ \bold{ \rightarrow{ \sqrt{ \dfrac{1 - sin \theta}{1 + sin \theta} } = sec \theta - tan \theta}}}$

$\dag{ \underline{ \underline{ \sf{Formulas \: applied}}}}$

$\\ \sf{ \rightarrow{(a + b)(a - b) = {a}^{2} - {b}^{2} }}$

$\sf{ \rightarrow{1 - {sin}^{2} \theta = {cos}^{2} \theta}}$

$\dag{ \underline{ \underline{ \sf{Solution}}}}$

L. H. S. =>

$\\ \rm{ \bold{ \sqrt{ \dfrac{1 - sin \theta}{1 + sin \theta}}}}$

$\\ \sf{ \implies{ \sqrt{ \frac{(1 - sin \theta)(1 - sin \theta)}{(1 + sin \theta)(1 + sin \theta)}}}}$

$\\ \sf{ \implies{ \sqrt{ \dfrac{(1 - sin \theta) {}^{2} }{ {1}^{2} - {sin}^{2} \theta}}}}$

$\\ \sf{ \implies{ \sqrt{ \dfrac{(1 - sin \theta) {}^{2} }{ {cos}^{2} \theta}}}}$

$\\ \sf{ \implies{ \frac{1 - sin \theta}{cos \theta}}}$

$\\ \sf{ \implies{ \frac{1}{cos \theta} - \frac{sin \theta}{cos \theta}}}$

$\\ \sf{ \implies{ \bold{ \red{sec \theta - tan \theta}}}} \: \: \: \: \: \: \: \boxed{ \rm{ \because{ \frac{1}{cos \theta = sec \theta} \: and \: \frac{sin \theta}{cos \theta} } = tan \theta}}$

=> R. H. S.

Hence, proved!!

$\dag{ \underline{ \underline{ \sf{Question \: (2)}}}}$

$\\\rm{ \bold{ \rightarrow{ \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } = cot \theta + cosec \theta}}}$

$\dag{ \underline{ \underline{ \sf{Formulas \: applied}}}}$

$\\ \sf{ \rightarrow{(a + b)(a - b) = {a}^{2} - {b}^{2} }}$

$\sf{ \rightarrow{ {sec}^{2} \theta + 1 = {tan}^{2} \theta}}$

$\dag{ \underline{ \underline{ \sf{Solution}}}}$

L. H. S. =>

$\\ \rm{ \bold{ \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1}}}}$

$\\ \sf{ \implies{ \sqrt{ \frac{(sec \theta + 1)(sec \theta + 1)}{(sec \theta - 1)(sec \theta + 1)}}}}$

$\\ \sf{ \implies{ \sqrt{ \frac{(sec \theta + 1) {}^{2} }{ {sec}^{2} \theta + 1}}}}$

$\\ \sf{ \implies{ \sqrt{ \frac{(sec \theta + 1) {}^{2} }{ {tan}^{2} \theta }}}}$

$\\ \sf{ \implies{ \frac{sec \theta + 1}{tan \theta}}}$

$\\ \sf{ \implies{ \frac{sec \theta}{tan \theta} + \frac{1}{tan \theta}}}$

$\\ \sf{ \implies{ \dfrac{ \dfrac{1}{cos \theta} }{ \dfrac{sin \theta}{cos \theta} } + cot \theta}} \: \: \: \: \: \: \: \boxed{ \because{ \rm{sec \theta = \frac{1}{cos \theta} \: and \: \frac{1}{tan \theta} = cot \theta }}}$

$\\ \sf{ \implies{ \frac{1}{cos \theta} \times \frac{cos \theta}{sin \theta} + cot \theta}}$

$\\ \sf{ \implies{ \bold{ \red{cosec \theta + cot \theta}}}} \: \: \: \: \: \: \: \boxed{ \because{ \rm{ \frac{1}{sin \theta} = cosec \theta }}}$

=> R. H. S.

Hence, proved!