Question

Prove the following trigonometric identities:
(1+tan²θ)(1-sinθ)(1+sinθ)=1

Answer 1

QuesTion :

$\normalsize\sf\ (1 + tan^2\theta)(1 - sin\theta)(1+sin\theta) = 1$

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AnswEr :

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: the \: question:}}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{Taking \: L.H.S}) }$

$\normalsize\hookrightarrow\sf\ (1 + tan^2\theta) \underbrace{(1 - sin\theta)(1 + sin\theta)}_{\orange{Multiplying \: both} \: }$

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

$\normalsize\hookrightarrow\sf\ (1 + Tan^2\theta)(1 - sin^2\theta)$

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

$\normalsize\hookrightarrow\sf\ (sec^2\theta)(cos^2\theta)$

$\scriptsize\sf{\: \: \: \: \: \:[\therefore\ \: \pink{ sec\theta = \frac{1}{cos\theta}}] }$

$\normalsize\hookrightarrow\sf\frac{1}{\cancel{cos^2\theta}} \times\ \cancel{cos^2\theta} \\ \\ \normalsize\hookrightarrow\sf\ 1 \: = \: R.H.S \\ \\ \normalsize\hookrightarrow\sf\ L.H.S \: = \: R.H.S$

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

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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

Answer:

1

Step-by-step explanation:

Given :

${\sf{\ \ (1 + tan^2 \theta )(1 - sin \theta )(1 + sin \theta )}}$

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${\boxed{\sf{\red{Identity \ : \ 1 + tan^2 \theta = sec^2 \theta }}}}$

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$\Rightarrow{\sf{(sec^2 \theta )(1 - sin \theta)(1 + sin \theta)}}$

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${\boxed{\sf{\red{Identity \ : \ (a - b)(a + b) = a^2 - b^2}}}}$

${\sf{\red{Here, \ a = 1, \ b = sin \theta}}}$

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$\Rightarrow{\sf{ (sec^2 \theta) [ (1)^2 - (sin \theta)^2 ] }}$

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$\Rightarrow{\sf{(sec^2 \theta)(1 - sin^2 \theta)}}$

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${\boxed{\sf{\red{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}}$

${\sf{\red{From \ this, \ we \ get \ [cos^2 \theta = 1 - sin^2 \theta] }}}$

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$\Rightarrow{\sf{(sec^2 \theta)(cos^2 \theta)}}$

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${\boxed{\sf{\red{Identity \ : \ sec^2 \theta = {\dfrac{1}{cos^2 \theta}} }}}}$

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$\Rightarrow{\sf{ \left( {\dfrac{1}{cos^2 \theta}} \right) (cos^2 \theta)}}$

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$\Rightarrow{\sf{ {\dfrac{1}{cos^2 \theta}} \times cos^2 \theta}}$

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$\Rightarrow{\sf{ {\dfrac{1}{{\cancel{cos^2 \theta}}}} \times {\cancel{cos^2 \theta}}}}$

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$\Rightarrow{\boxed{\sf{\green{1}}}}$