Question

Prove the following trigonometric identities:
1-sinθ/1+sinθ=(secθ-tanθ)²

Answer 1

QuesTion :

$\large\sf\frac{1 - sin\theta}{1 + sin\theta} = (sec\theta + tan\theta)^2$

AnswEr :

$\scriptsize\sf{\: \: \: \: \: \:[ \therefore\ \: \pink{Taking \: R.H.S}]}$

$\normalsize\hookrightarrow\sf\ (sec\theta - tan\theta)^2$

$\scriptsize\sf{\: \: \: \: \: \:[ \therefore\ \: \pink{ sec\theta = \frac{1}{cos\theta} \: and \: tan\theta = \frac{sin\theta}{cos\theta} }) }$

$\normalsize\hookrightarrow\sf\ [\frac{1}{cos\theta} + \frac{cos\theta}{sin\theta}]^2 \\ \\ \normalsize\hookrightarrow\sf\ [\frac{1 - sin\theta}{cos\theta}]^2$

$\scriptsize\sf{\: \: \: \: \: \:[ \therefore\ \: \pink{ cos^2\theta = 1 - sin^2\theta}] }$

$\normalsize\hookrightarrow\sf\frac{(1-sin\theta)(1+sin\theta)}{1 - sin^2\theta} \\ \\ \normalsize\hookrightarrow\sf\frac{\cancel{(1 - sin\theta)}(1+sin\theta)}{(1-sin\theta) \cancel{(1-sin\theta)} }$

$\normalsize\hookrightarrow\sf\frac{1+sin\theta}{1-sin\theta} \\ \\ \normalsize\hookrightarrow\sf\ L.H.S \\ \\ \normalsize\hookrightarrow\sf\ L.H.S = R.H.S$

$\LARGE\hookrightarrow{\underline{\boxed{\sf\red{Hence \: proved!! }}}}$

$\rule{100}2$

Some important identities related to it :

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

Answer 2

★$\mathbb{QUESTION}$★

$\frac{1-sin\theta}{1+sin\theta}$${=}$${(sec\theta-tan)^2}$

★$\mathbb{SOLUTION}$★

LHS

→$\frac{1-sin\theta}{1+sin\theta}$

→$\large\frac{1}{cos\theta}$${-}$$\large\frac{sin\theta}{cos\theta}$${÷}$$\large\frac{1}{cos\theta}$${+}$$\large\frac{sin\theta}{cos\theta}$

→$\frac{sec\theta-tan\theta}{sec\theta+tan\theta}$

→$\frac{sec\theta-tan\theta}{sec\theta+tan\theta}×\frac{sec\theta-tan\theta}{sec\theta-tan\theta}$

→$\frac{(sec\theta-tan\theta)^2}{sec^2\theta-tan^2\theta}$

→${(sec\theta-tan\theta)^2}$

= RHS verified

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$\huge\underline\frak\red{note}$

• Divide both numerator and denominator by cos∅

• 1/cos∅ = sec∅

• sin∅/cos∅ = tan∅

• Rationalize

• sec²∅ - tan²∅ = 1

•(a+b)(a-b) = a²-b²

[used in this proven]