Question

1 + sec theta/sec theta=sin^2 theta/1-cos theta. Prove that
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Answer 1

Topic :-

Trigonometry

To Prove :-

$\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}$

Proof :-

Solving LHS,

$\leadsto\dfrac{1+\sec\theta}{\sec\theta}$

$\leadsto\dfrac{1}{\sec\theta}+\dfrac{\sec\theta}{\sec\theta}$

$\leadsto\dfrac{1}{\sec\theta}+1$

$\leadsto\cos\theta+1$

$\left(\because\dfrac{1}{\sec\theta}=\cos\theta \right)$

Solving RHS,

$\leadsto\dfrac{\sin^2\theta}{1-\cos\theta}$

$\leadsto\dfrac{1-\cos^2\theta}{1-\cos\theta}$

$\left(\because\sin^2\theta+\cos^2\theta=1 \right)$

$\leadsto\dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}$

$\left(\because a^2-b^2=(a-b)(a+b) \right)$

$\leadsto\dfrac{\cancel{1-\cos\theta}}{\cancel{1-\cos\theta}}\cdot(1+\cos\theta)$

$\leadsto 1+\cos\theta$

We observe that LHS = RHS.

Hence, Proved !!

Additional Formulae :-

$1+\tan^2\theta=\sec^2\theta$

$1+\cot^2\theta=\csc^2\theta$

$1+\cos2\theta=2\cos^2\theta$

$1-\cos2\theta=2\sin^2\theta$

$\sin2\theta=2\sin\theta\cos\theta$

$\cos2\theta=\cos^2\theta-\sin^2\theta$

$\sin3\theta=3\sin\theta-4\sin^3\theta$

$\cos3\theta=4\cos^3\theta-3\cos\theta$

$(\sin\theta+\cos\theta)^2=1+\sin2\theta$

$(\sin\theta-\cos\theta)^2=1-\sin2\theta$