Question

prove the identity (1-cos^2 theta)sec^2 theta = tan^2 theta​

Answer 1

Answer:

Lhs=Rhs

Step-by-step explanation:

Explanation:

sec2θ(1−cos2θ)=tan2θ

Left Side :=sec2θ(1−cos2θ)

=sec2θsin2θ

=1-cos2θ⋅sin2θ

=sin2θcos2θ

=tan2θ

∴= Right Side

Answer 2

$\large{\underline{\bf{\green{To\:prove:-}}}}$

(1-cos^2 theta)sec^2 theta = tan^2 theta

$\huge{\underline{\bf{\red{Solution:-}}}}$

$: \implies \bf\:(1 - {cos}^{2} \theta) {sec}^{2} \theta = {tan}^{2} \theta \\ \\: \implies \bf\:LHS \\ \\ : \implies \bf\:(1 - {cos}^{2} \theta) \frac{1}{ {cos}^{2} \theta} \\ \\ : \implies \: \: { \boxed{ \purple{ \bf{ {sin}^{2} \theta = 1 - {cos}^{2} \theta }}}}\\ \\: \implies \bf\: ({sin}^{2} \theta) \frac{1}{cos {}^{2} \theta } \\ \\ : \implies \: { \boxed{ \bf{ \purple{ \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta } = {tan}^{2} \theta }}}} \\ \\: \implies \: \bf\frac{ {sin}^{2} \theta }{ {cos}^{2} \theta} \\ \\ : \implies\bf \: \bf {tan}^{2} \theta$

Hence proved

LHS = RHS

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Some important identities:-

${ \blue{ \bf{cosec\theta=\frac{1}{sin\theta}}}}$

${ \blue{ \bf{sec\theta=\frac{1}{cos\theta}}}}$

${ \blue{ \bf{cot\theta=\frac{1}{tan\theta}}}}$

${ \blue{ \bf{tan\theta=\frac{sin\theta}{cos\theta}}}}$

${ \blue{ \bf{cot\theta=\frac{cos\theta}{sin\theta}}}}$

${ \blue{ \bf{tan\theta.cot\theta=1}}}$

${ \blue{ \bf{sin^2\theta+cos^2\theta=1}}}$

${ \blue{ \bf{1+tan^2\theta=sec^2\theta}}}$

${ \blue{ \bf{1+cot^2\theta=cosec^2\theta}}}$

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