Question

prove that, sec θ (1 − sin θ) (sec θ + tan θ) = 1

Answer 1

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Answer 2

TO PROVE :–

$\\ \to \: { \bold{ \sec( \theta) \{1 - \sin( \theta) \} \{ \sec( \theta) + \tan( \theta) \} = 1}}$

SOLUTION :–

• Let's take L.H.S. –

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

• We know that –

$\\ \: \: \blacktriangleright \: \: { \bold{ \sec( \theta) = \dfrac{1}{ \cos( \theta) } \: \: and \: \: \tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) } }}$

• So that –

$\\ \: = \: { \bold{ \dfrac{1}{\cos( \theta)} \{1 - \sin( \theta) \} \left\{ \dfrac{1}{\cos( \theta)} + \dfrac{ \sin( \theta) }{ \cos( \theta) } \right \} }}$

$\\ \: = \: { \bold{ \dfrac{1}{\cos( \theta)} \{1 - \sin( \theta) \} \left\{ \dfrac{ 1 + \sin( \theta) }{ \cos( \theta) } \right \} }}$

$\\ \: = \: { \bold{ \dfrac{1}{\cos( \theta)} \left\{ \dfrac{ \{1 - \sin( \theta) \} \{1 + \sin( \theta) \}}{ \cos( \theta) } \right \} }}$

$\\ \: = \: { \bold{ \dfrac{ \{1 - \sin( \theta) \} \{1 + \sin( \theta) \}}{ \cos^{2} ( \theta) } }}$

$\\ \: = \: { \bold{ \dfrac{1 - \sin^{2} ( \theta)}{ \cos^{2} ( \theta) } \: \: \: \: \: \left[ \: \because \: (a - b)(a + b) = {a}^{2} - {b}^{2} \right]}}$

$\\ \: = \: { \bold{ \cancel \dfrac{{ \cos}^{2}( \theta)}{ \cos^{2} ( \theta) } \: \: \: \: \: \left[ \: \because \: 1 - \sin^{2} ( \theta) = { \cos}^{2}( \theta) \right]}}$

$\\ { \bold{ \: \: = \: \: 1}}$

$\\ { \bold{ \: \: = \: \: R.H.S.}}$

$\\ { \bold{ \: \: \: \: \: \: \: \: \: \: \underbrace{ Hence \: \: proved}}}$