Questions & Answers
Question
AnswersIf ${\tan ^2}\theta = 2{\tan ^2}\phi + 1$ . Then find the value of $\cos 2\theta + {\sin ^2}\phi .$
A) -1
B) 0
C) 1
D) None of these
Answer
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Hint:To solve the problem we have to find the value of $\cos 2\theta $ in terms of $\phi $ . Then we have to calculate the value of $\cos 2\theta + {\sin ^2}\phi .$
Complete step-by-step answer:
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).
Note: $\sin \theta $ , $\cos \theta $ , $\tan \theta $ etc. are called circular angles. These are the ratios of three sides of a right angle triangle taking two sides at a time. The ratio of perpendicular and hypotenuse is $\sin \theta $, the ratio of base and hypotenuse is $\cos \theta $ and the ratio of perpendicular and base is $\tan \theta $. The reciprocal of $\sin \theta $ is $\co sec\theta $ , the reciprocal of $\cos \theta $ is $\sec \theta $ and the reciprocal of $\tan \theta $ is $\cot \theta .$ Also $\tan \theta $ is the ratio of $\sin \theta $ and $\cos \theta $, $\cot \theta $ is the ratio of $\cos \theta $ and $\sin \theta $ . If an angle be $\theta $ , then $2\theta $ is the multiple angle. So $\sin 2\theta ,\cos 2\theta $ are the sine and cosine formulae for multiple angles.
Complete step-by-step answer:
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).
Note: $\sin \theta $ , $\cos \theta $ , $\tan \theta $ etc. are called circular angles. These are the ratios of three sides of a right angle triangle taking two sides at a time. The ratio of perpendicular and hypotenuse is $\sin \theta $, the ratio of base and hypotenuse is $\cos \theta $ and the ratio of perpendicular and base is $\tan \theta $. The reciprocal of $\sin \theta $ is $\co sec\theta $ , the reciprocal of $\cos \theta $ is $\sec \theta $ and the reciprocal of $\tan \theta $ is $\cot \theta .$ Also $\tan \theta $ is the ratio of $\sin \theta $ and $\cos \theta $, $\cot \theta $ is the ratio of $\cos \theta $ and $\sin \theta $ . If an angle be $\theta $ , then $2\theta $ is the multiple angle. So $\sin 2\theta ,\cos 2\theta $ are the sine and cosine formulae for multiple angles.