Question

Simplify : (1 + tan2 theta)(1 - sin theta) (1 + sin theta)​

Answer 1

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

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

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

Answer 2

$\bf Given:\ \sf (1\ +\ tan^2\ \theta)(1\ -\ sin\ \theta)(1\ +\ sin\ \theta).\\\\\bf To\ simplify\ it:\\\\\\\sf (1\ +\ tan^2\ \theta)(1\ -\ sin\ \theta)(1\ +\ sin\ \theta)\\\\\\\tt We\ know\ that\ sec^2\ \theta\ -\ tan^2\ \theta\ =\ 1.\\\\Hence, sec^2\ \theta\ =\ 1\ +\ tan^2\ \theta.\\\\\\\sf (sec^2\ \theta)(1\ -\ sin\ \theta)(1\ +\ sin\ \theta)\\\\\\\tt We\ \know\ that\ (a\ +\ b)(a\ -\ b)\ =\ a^2\ -\ b^2.\\\\Hence, (1\ -\ sin\ \theta)(1\ +\ sin\ \theta)\ =\ 1\ -\ sin^2\ \theta.$

$\sf (sec^2\ \theta)(1\ -\ sin^2\ \theta)\\\\\\\tt We\ know\ that\ sin^2\ \theta\ +\ cos^2\ \theta\ =\ 1.\\\\Hence,\ cos^2\ \theta\ =\ 1\ -\ sin^2\ \theta.\\\\\\\sf (sec^2\ \theta)(cos^2\ \theta)\\\\\\\tt We\ know\ that\ sec^2\ \theta\ =\ \dfrac{1}{cos^2\ \theta}.\\\\\\\sf \bigg(\dfrac{1}{cos^2\ \theta}\bigg)\ \times\ \bigg(cos^2\ \theta\bigg)\\\\\\\bf =\ 1.$