Question

$\text\pink{\underline{\underline{Question}}}$
Prove the identities:
(i) √[1 + sinA/1 – sinA] = sec A + tan A​

Answer 1

$\bigstar\large \purple{\underline{ \pmb{ \sf to \: prove : - }}}$

$\sf \sqrt{ \dfrac{1 + \sin A}{1 - \sin A } } = \sec A + \tan A$

LHS :-

Multiply the L.H.S with the conjugates of 1+sinA

$: \implies\large\tt \sqrt{ \bigg(\dfrac{1 + \sin A}{1 - \sin A} \bigg) \bigg( \dfrac{1 + \sin A}{1 + \sin A} \bigg) }$

$: \implies\large \tt \sqrt{ \dfrac{(1 + \sin A) ^{2} }{1 - { \sin}^{2}A } }$

$\dag \bf by \: using : -$

• $\underline{ \boxed{ \sf 1 - { \sin}^{2}x = { \:cos}^{2}x }}$

$: \implies \large\tt \sqrt{ \dfrac{(1 + \sin A) ^{2} }{ { \cos}^{2}A } } = \dfrac{1 + \sin A}{ \cos A}$

$: \implies \large \tt\dfrac{1}{ \cos A} + \dfrac{ \sin A}{ \cos A} = \bf \red{ \sec A + \tan A}$

LHS = RHS

$\sf \pink{hence \: verified}$

ADDITIONAL KNOWLEDGE:-

Important Identities:-

• sec²∅ - tan²∅ = 1
• cosec²∅ - cot²∅ = 1

Reciprocal of ratios:-

• sin∅ = 1/cosec∅
• cosec∅ = 1/sin∅
• cos∅ = 1/sec∅
• sec∅ = 1/cos∅
• tan∅ = 1/cot∅
• cot∅ = 1/tan∅

Ratios of Complemantary Angles:-

• sin∅ = cos(90 - ∅)
• cos∅ = sin(90 - ∅)
• tan∅ = cot(90 - ∅)
• cot∅ = tan(90 - ∅)
• sec∅ = cosec(90 - ∅)
• cosec∅ = sec(90 = ∅)

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

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I hope it's beneficial :D