Question

Sec²theta - Sin²theta/ tan²theta = 1 +cot²theta - cos²theta​

Answer 1

ANSWER:

To Prove:

• (sec²θ - sin²θ)/(tan²θ) = 1 + cot²θ - cos²θ

Proof:

We are given that,

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

Taking LHS,

$\implies \dfrac{\sec^2\theta-\sin^2\theta}{\tan^2\theta}$

We know that,

$\hookrightarrow\sec^2\phi=\dfrac{1}{\cos^2\phi}$

So,

$\implies \dfrac{\sec^2\theta-\sin^2\theta}{\tan^2\theta}$

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

Taking LCM,

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

We know that,

$\hookrightarrow\tan^2\phi=\dfrac{\sin^2\phi}{\cos^2\phi}$

So,

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

Cancelling cos²θ,

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

Separating the terms,

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

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

We know that,

$\hookrightarrow \dfrac{1}{\sin^2\phi}=\csc^2\phi$

So,

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

$\implies \csc^2\theta-\dfrac{\sin^2\theta\cos^2\theta}{\sin^2\theta}$

Cancelling sin²θ,

$\implies \csc^2\theta-\cos^2\theta$

We know that,

$\hookrightarrow\csc^2\phi=1+\cot^2\phi$

So,

$\implies \csc^2\theta-\cos^2\theta$

$\implies 1+\cot^2\theta-\cos^2\theta$

$\implies\bf \dfrac{sec^2\theta-sin^2\theta}{tan^2\theta}=1+cot^2\theta-cos^2\theta$

As, LHS = RHS,

HENCE PROVED!!

Answer 2

Answer:

we have,

$\tt \dfrac{sec^2 \: \theta -sin^2 \: \theta}{tan^2 \: \theta}$

$\tt \implies \dfrac{\dfrac{1}{cos^2\: \theta} \: -sin^2 \: \theta}{tan^2 \: \theta}$

$\tt \implies \dfrac{\dfrac{1- sin^2 \: \theta cos^2 \: \theta}{cos^2\: \theta} }{tan^2 \: \theta}$

$\tt \implies \dfrac{\dfrac{1- sin^2 \: \theta cos^2 \: \theta}{cos^2\: \theta} }{\dfrac{sin^2 \: \theta}{cos^2 \: \theta}}$

$\tt \implies \dfrac{\dfrac{1- sin^2 \: \theta cos^2 \: \theta}{\cancel{cos^2\: \theta}}}{\dfrac{sin^2 \: \theta}{\cancel{cos^2 \: \theta}}}$

$\tt \implies \dfrac{1-sin^2 \: \theta cos^2 \: \theta}{sin^2 \: \theta}$

$\tt \implies \dfrac{1}{sin^2 \: \theta} - \dfrac{sin^2 \: \theta cos^2 \: \theta}{sin^2 \: \theta}$

$\tt \implies cosec^2 \: \theta - \dfrac{\cancel{sin^2 \: \theta} cos^2 \: \theta}{\cancel{sin^2 \: \theta}}$

$\tt \implies cosec^2 \: \theta - cos^2 \: \theta$

$\tt \implies 1+ cot^2 \: \theta - cos^2 \: \theta$