Question

sec∅ + tan∅ + 1 / sec ∅ - tan ∅ + 1 = 1 + sin ∅/cos ∅.

Solve this question ....​

Answer 1

Question :–

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

ANSWER :–

TO PROVE :–

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

SOLUTION :–

• Let's take L.H.S. –

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

• We know that –

$\\ \: \longrightarrow \: { \bold{ \: \sec {}^{2} ( \theta) - \tan {}^{2} ( \theta) = 1 }} \: \:$

• So that –

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

• Using identity –

$\\ \: \longrightarrow \: { \bold{ \: {a}^{2} - {b}^{2} = (a + b)(a - b) }} \: \:$

$\\ \: \: { \bold{ \: = \: \dfrac{ \sec( \theta) + \tan( \theta) + [ { \sec}^{} ( \theta) - { \tan}^{}( \theta ) ][ \sec( \theta) + \tan( \theta) ] }{\sec( \theta) - \tan( \theta) + 1} }} \: \:$

$\\ \: \: { \bold{ \: = \: \dfrac{ [ \sec( \theta) + \tan( \theta) ] { \cancel{[ { \sec}^{} ( \theta) - { \tan}^{}( \theta ) + 1 ]}}}{ \cancel{ [ \sec( \theta) - \tan( \theta) + 1 ]}} }} \: \:$

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

• We know that –

$\\ \: \longrightarrow \: { \bold{ \: \sec ( \theta) = \dfrac{1}{ \cos( \theta) } }} \: \:$

$\\ \: \longrightarrow \: { \bold{ \: \tan ( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) } }} \: \:$

• So that –

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

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

$\\ \: \: { \bold{ \: = R.H.S. \: \: \: \: \: \: \: \: (Hence \: \: proved)}} \: \:$

Answer 2

$\large\bf\underline{Question:-}$

Prove that :- sec∅ + tan∅ + 1 / sec ∅ - tan ∅ + 1 = 1 + sin ∅/cos ∅.

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$\large\bf\underline{Given:-}$

LHS = sec∅ + tan∅ + 1 / sec ∅ - tan ∅ + 1

RHS = 1 + sin ∅/cos ∅.

$\large\bf\underline {To \: prove:-}$

sec∅ + tan∅ + 1 / sec ∅ - tan ∅ + 1 = 1 + sin ∅/cos ∅.

$\huge\bf\underline{Solution:-}$

$: \implies \rm \: \frac{sec \theta +t an \theta + 1}{sec \theta - tan \theta + 1} = \frac{1 + sin \theta}{cos \theta}$

LHS:-

We know that,

$\bf \: 1 = {sec}^{2} \theta - {tan}^{2 } \theta$

Then,

$: \implies \rm \: \frac{sec \theta + tan \theta + (se {c}^{2} \theta - {tan}^{2} \theta) }{sec \theta - tan \theta + 1}$

By using identity :

$\bf \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$: \implies \rm \: \frac{sec \theta + tan \theta + (sec \theta + tan \theta)(sec \theta - tan \theta}{sec \theta - tan \theta + 1)} \\ \\ : \implies \rm \: \frac{(sec \theta + tan \theta) \cancel{(1 + (sec \theta - tan \theta)}}{{ \cancel{sec \theta - tan \theta + 1}}} \\ \\ : \implies \rm \: sec \theta + tan \theta$

we know that,

$\bf \: sec \theta = \frac{1}{cos \theta} \\ \\ \bf \:tan \theta = \frac{sin \theta}{ \cos \theta}$

$: \implies \rm \: \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta} \\ \\ : \implies \rm \: \frac{1 + sin \theta}{cos \theta} = RHS$

LHS = RHS

hence ,proved

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