Questions & Answers

Question

Answers

A) -1

B) 0

C) 1

D) None of these

Answer

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From the given equation we have ${\tan ^2}\theta = 2{\tan ^2}\phi + 1$.

We know that $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$ .

Let us substitute the value of ${\tan ^2}\theta $ in the above formula, we get,

$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \dfrac{{1 - (2{{\tan }^2}\phi + 1)}}{{1 + (2{{\tan }^2}\phi + 1)}}.$

Now we can simplify the above equation to get the following equation, $\cos 2\theta = \dfrac{{ - 2{{\tan }^2}\phi }}{{2 + 2{{\tan }^2}\phi }}.$

Taking 2 common from denominator, $\cos 2\theta = \dfrac{{ - 2{{\tan }^2}\phi }}{{2(1 + {{\tan }^2}\phi )}}.$

Cancelling the common factor 2 from both numerator and denominator, we get,

$\cos 2\theta = \dfrac{{ - {{\tan }^2}\phi }}{{1 + {{\tan }^2}\phi }}.$

We know that ${\tan ^2}\phi = \dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }}$ therefore, we get,

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

Further simplifying the above equation we get, $\cos 2\theta = - \dfrac{{\dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }}}}{{\dfrac{{{{\cos }^2}\phi + {{\sin }^2}\phi }}{{{{\cos }^2}\phi }}}}$

We now that ${\cos ^2}\phi + {\sin ^2}\phi = 1$substitute the identity in the above equation, we get,

\[\cos 2\theta = - \dfrac{{\dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }}}}{{\dfrac{1}{{{{\cos }^2}\phi }}}}\]

Further simplifying the values we get,

\[\cos 2\theta = - \dfrac{{{{\sin }^2}\phi }}{{{{\cos }^2}\phi }} \times \dfrac{{{{\cos }^2}\phi }}{1} = - {\sin ^2}\phi .\]

Now we calculate the value of $\cos 2\theta + {\sin ^2}\phi .$

Putting the value of$\cos 2\theta = - {\sin ^2}\phi $, we obtain,

$\cos 2\theta + {\sin ^2}\phi = - {\sin ^2}\phi + {\sin ^2}\phi = 0.$

$\cos 2\theta + {\sin ^2}\phi $= 0

Hence we have come to the conclusion that the correct answer is option (B).