Question

If $\theta $ is an acute angle such that ${{\tan }^{2}}\theta =\dfrac{8}{7}$, then the value of $\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$ is

a)$\dfrac{7}{8}$
b)$\dfrac{8}{7}$
c)$\dfrac{7}{4}$
d)$\dfrac{64}{49}$

Answer 1

Hint: Here, first we have to apply the formula $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$, since the numerator and denominator of the expression $\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$ is of the form (a + b)(a – b). After this, we have to use the identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and then we can find the value of ${{\cot }^{2}}\theta $ by the formula $\cot \theta =\dfrac{1}{\tan \theta }$.

Complete step-by-step answer:
Here, we are given that $\theta $ is an acute angle such that ${{\tan }^{2}}\theta =\dfrac{8}{7}$.

Now, we have to find the value of $\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$.

First let us consider the expression $\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$. It is of the form (a + b)(a – b).

We know that,

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

By applying this property we obtain,

$\Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{{{1}^{2}}-{{\sin }^{2}}\theta }{{{1}^{2}}-{{\cos }^{2}}\theta }$

$\Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }$ ……. (1)

We have the identity,

${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

Now, by taking ${{\sin }^{2}}\theta $ to the right side we get,

$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ ……. (2)

Similarly, by taking ${{\cos }^{2}}\theta $ to the right side, we get,

$\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
                                                 …….. (3)

Now, by substituting equation (2) and equation (3) in equation (2) we obtain,

$\Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta

)}=\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$

We also have the formula that,

$\begin{align}

  & \cot \theta =\dfrac{\cos \theta }{\sin \theta } \\

 & \Rightarrow {{\cot }^{2}}\theta =\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \\

\end{align}$

 Therefore, we can write:

$\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}={{\cot }^{2}}\theta $

We also know that,

$\begin{align}

  & \cot \theta =\dfrac{1}{\tan \theta } \\

 & \Rightarrow {{\cot }^{2}}\theta =\dfrac{1}{{{\tan }^{2}}\theta } \\

\end{align}$

Hence, we will get,

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

Now, by substituting the value of ${{\tan }^{2}}\theta =\dfrac{8}{7}$ w get:

$\begin{align}

  & \Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{1}{\dfrac{8}{7}} \\

 & \Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=1\times \dfrac{7}{8} \\

 & \Rightarrow \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{7}{8} \\

\end{align}$

Hence, we can say that when $\theta $ is an acute angle such that ${{\tan }^{2}}\theta =\dfrac{8}{7}$, then the value of $\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{7}{8}$.

Therefore, the correct answer for this question is option (a).


Note: Alternate Solution: Here, it is given as ${{\tan }^{2}}\theta =\dfrac{8}{7}$, now by taking the square root we will get the value of $\tan \theta $. Then, apply the formula $\tan \theta =\dfrac{Opposite\text{ }side}{Adjacent\text{ }side}\text{ }$, where opposite side and adjacent side are given. Next, find the value of hypotenuse with the help of Pythagora's theorem which is then used to find $\sin \theta $ and $\cos \theta $.