Question

prove (1+ tan^ 2 theta )( 1+ sin theta)( 1- sin theta) is equal to 1

Answer 1

Answer:

We have to show that:

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

Let us start by taking left hand side:

$L.H.S=(1+\tan^2\theta)(1+\sin\theta)(1-\sin\theta)$

As we know:-

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

and

$(a+b)(a-b)=a^2-b^2$

$L.H.S=(\sec^2\theta)((1)^2-(\sin\theta)^2)\\L.H.S=(\sec^2\theta)(1-\sin^2\theta)$

We know that:-

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

Finally we have,

$L.H.S.=\dfrac{1}{\cos^2\theta}\times \cos^2\theta=1\\L.H.S=R.H.S$

Answer 2

Given that,

The function is

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

We need to prove the Left hand side equal to right hand side

Using given function

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

Using left hand side,

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

We know that,

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

Now put the value of  $[tex](1+\tan^2\theta)$

$=\sec^2\theta(1-\sin^2\theta)$

We know that,

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

$=\sec^2\theta\cos^2\theta$

$=\sec^2\theta\times\dfrac{1}{\sec^2\theta}$

$=1$

R.H.S

Hence, That is proved.