Question

Question : if B=x, then prove that $(x^2+1)cos\theta+(x^2+1)sin\theta=(x+1)^2-2$​

Answer 1

Given that :

• $\sf{B=tan\theta+sec\theta}$

According to question, B = x

So, $\tt{tan\theta+sec\theta=x}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{sin\theta}{cos\theta}+\frac{1}{cos\theta}}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{sin\theta+1}{cos\theta}=x}}}\end{gathered}$

By squaring bothe side :

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{(1+sin\theta)^2}{cos^2\theta}=x^2}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{(1+sin\theta)^2}{1+sin^2\theta}=x^2}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{1+sin\theta}{1-sin\theta}=x^2}}}\end{gathered}$

By componendo - dividendo :

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{1+sin\theta-1+sin\theta}{1+sin\theta+1-sin\theta}+\frac{x^2-1}{x^2+1}}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\longmapsto{\tt{\frac{2sin\theta}{2}=\frac{x^2-1}{x^2+1}}}}\end{gathered}$

$\begin{gathered}\:\displaystyle{\tt{sin\theta=\frac{x^2-1}{x^2+1}}}\end{gathered}$

____________________________

$\sf{cos\theta=\sqrt{1-sin^2\theta}}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{=\sqrt{1-{\frac{x^2-1}{x^2+1}}^2}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{=\sqrt{1-\frac{x^4-2x^2+1}{x^4+2x^2+1}}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{=\sqrt{\frac{x^4-2x^2+1-x^4+2x^2-1}{x^4+2x^2+1}}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{=\sqrt{\frac{4x^2}{(x^2+1)^2}}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{=\frac{2x}{x^2+1}}}\end{gathered}$

_______________________________

$\sf{L.H.S}\tt{= ( {x}^{2} + 1) \cos\theta + ( {x}^{2} + 1) \sin\theta}$

$\:\:\:\:\:\:\:\:\:\:\tt{= ( {x}^{2} + 1)( \cos\theta + \sin\theta)}$

$\:\:\:\:\:\:\:\:\:\:\tt{= ( {x}^{2} + 1)( \frac{2x}{ {x}^{2} + 1} + \frac{ {x}^{2} - 1 }{ {x}^{2} + 1 } )}$

$\:\:\:\:\:\:\:\:\:\:\tt{= ( {x}^{2} + 1)( \frac{2x + {x}^{2} - 1 }{ {x}^{2} + 1 } )}$

$\:\:\:\:\:\:\:\:\:\:\tt{= {x}^{2} + 2x - 1}$

$\:\:\:\:\:\:\:\:\:\:\tt{= {x}^{2} + 2x + 1 - 2}$

$\:\:\:\:\:\:\:\:\:\:\tt{= {(x + 1)}^{2} - 2}$

$\sf{\:\:\:\:\:\:\:\:\:\:=R.H.S}$

$\begin{gathered}\:\:\underline{\boxed{\tt{Hence,( {x}^{2} + 1) \cos\theta + ( {x}^{2} + 1)=(x + 1)^2 - 2}}}\end{gathered}$

Answer 2

Answer:

refer to attachment

hope its helpful