Question

(1+tan^2theta) (1-sin theta) (1+sin theta)=1​

Answer 1

Answer:

We have to show that:

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

2

θ)(1+sinθ)(1−sinθ)=1

Let us start by taking left hand side:

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

2

θ)(1+sinθ)(1−sinθ)

As we know:-

1+\tan^2\theta=\sec^2\theta1+tan

2

θ=sec

2

θ

and

(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a

2

−b

2

\begin{gathered}L.H.S=(\sec^2\theta)((1)^2-(\sin\theta)^2)\\L.H.S=(\sec^2\theta)(1-\sin^2\theta)\end{gathered}

L.H.S=(sec

2

θ)((1)

2

−(sinθ)

2

)

L.H.S=(sec

2

θ)(1−sin

2

θ)

We know that:-

\begin{gathered}\sec^2\theta=\dfrac{1}{\cos^2\theta}\\1-\sin^2\theta=cos^2\theta\end{gathered}

sec

2

θ=

cos

2

θ

1

1−sin

2

θ=cos

2

θ

Finally we have,

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

L.H.S.=

cos

2

θ

1

×cos

2

θ=1

L.H.S=R.H.S