​$$\text { If } \cos ^4 \theta-\sin ^4 \theta=k \text {, then the value of } \frac{1+k}{1-k} \text { is: }$$​​

Correct option is A

Given:

$$\text { } \cos ^4 \theta-\sin ^4 \theta=k \text {, then } \frac{1+k}{1-k} \text {= ? }$$

Formula used:

$$\begin{aligned}& \cos ^2 \theta+\sin ^2 \theta=1 \\& \cot \theta=\frac{\cos \theta }{ \sin \theta}\end{aligned}$$

(a² - b²) = ( a +b) ( a - b) 

Solution:

$$\begin{aligned}& \cos ^4 \theta-\sin ^4 \theta=\mathrm{k} \\& \Rightarrow\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right)=\mathrm{k} \\& \Rightarrow 1 \times\left(\cos ^2 \theta-\sin ^2 \theta\right)=\mathrm{k} \\& \Rightarrow \mathrm{k}=\cos ^2 \theta-\sin ^2 \theta\end{aligned}$$

Now, we have to find the value of  $$\frac{(1 +k)}{(1 - k)}$$​

$$\begin{aligned}& \Rightarrow \frac{1+\left(\cos ^2 \theta-\sin ^2 \theta\right)}{1-\left(\cos ^2 \theta-\sin ^2 \theta\right)} \\& \Rightarrow \frac{\cos ^2 \theta+\sin ^2 \theta+\cos ^2 \theta-\sin ^2 \theta}{\cos ^2 \theta+\sin ^2 \theta-\cos ^2 \theta+\sin ^2 \theta}=\frac{2 \cos ^2 \theta }{ 2 \sin ^2 \theta} \\& \Rightarrow \frac{\cos ^2 \theta}{ \sin ^2 \theta}=\cot ^2 \theta\end{aligned}$$

Hence, the correct answer is option(a).

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