Question

Trignometry question Prove that $\sf\sqrt{1+\sin\:A}{1-\sin\:A}=\sec\:A+\tan\:A$ Quality answer needed.

Answer 1

Correct Question:-

Trignometry question

Prove that $\sf \sqrt{ \dfrac{ 1 + sin\;A}{1 - sin\;A}} = sec\;A + tan\;A$ Quality answer needed.

To prove:-

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

Proof:-

$\small\sf\;\;\dag\; \underline{Taking\;LHS:-}$

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$:\implies\sf \sqrt{ \dfrac{ 1 + sin\;A}{1 - sin\;A}}$

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$\small\sf\;\;\dag\; \underline{Rationalize\;the\; denominator:-}$

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$:\implies\sf \sqrt{ \dfrac{ 1 + sin\;A}{1 - sin\;A} \times \dfrac{ 1 + sin\;A}{1 + sin\;A}}$

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$:\implies\sf \sqrt{ \dfrac{ 1 + sin\;A}{1 - sin\;A} \; \dfrac{ 1 + sin\;A}{1 + sin\;A}}$

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✮ As we know the identity,

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${\underline{\boxed{\bf{\red{x^2 - y^2 = (x - y)(x + y)}}}}}$

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

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✮ Here, we use the Trigonometric identity -

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${\underline{\boxed{\bf{\purple{cos^2\;A = 1 - sin^2\;A}}}}}$

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$:\implies\sf \sqrt{ \bigg( \dfrac{1 + sin\;A}{cos\;A} \bigg) }$

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$:\implies\sf \dfrac{1 + sin\;A}{cos\;A}$

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$:\implies\sf \dfrac{1}{cos\;A} + \dfrac{sin\;A}{cos\;A}$

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As we know that,

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★ $\sf \dfrac{1}{cos\;A} = sec\;A$

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And

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★ $\sf \dfrac{sin\;A}{cos\;A} = tan\;A$

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Therefore,

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$:\implies\sf \dfrac{1}{cos\;A} + \dfrac{sin\;A}{cos\;A}$

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$:\implies{\underline{\boxed{\bf{\blue{sec\;A + tan\;A}}}}}$

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$\dag\sf \underline{Hence,\;LHS = RHS}$

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Additional Information:-

$\begin{lgathered}\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\cos^2\theta=1-\sin^2\theta \\ \\ 4)1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5) \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\sec^2\theta=1+\tan^2\theta \\ \\ 8)\sec^2\theta-\tan^2\thetha=1 \\ \\ 9)\tan^2\theta=\sec^2\theta-1 \end{minipage}}\end{lgathered}$

Answer 2

Correct Question :

Prove that

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

Solution :

We have to prove that

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

LHS

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

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

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

We know that sin² x+cos² x=1

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

$\sf=\dfrac{(1+\sin\:A)}{cosA}$

RHS

$\sf=\sec\:A+\tan\:A$

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

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

Now , LHS = RHS

Hence Proved

$\rule{200}2$

Trignometric Formulas :

sinA cos B + sinB cosA = sin( A+ B)

sinAcosB - sinBcosA =sin(A-B)

cosAcosB -sinAsinB=cos( A+B)

cosAcosB+sinAsinB =cos(A-B)

sin²A + cos²A = 1

sec²A - tan²A = 1

cosec²A - cot²A = 1

sin2A = 2 sinA cosA

cos2A = cos²A - sin²A

tan2A = 2 tanA / (1 - tan²A)