Question

If the value of $\tan \theta =\dfrac{1}{\sqrt{7}}$ , show that $\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$ .

Answer 1

Hint:In the given expression, we don’t have any $\tan \theta$ term. For finding the value of the given expression, we have to make the expression in the form of $\tan \theta$. For that take ${{\operatorname{cosec}}^{2}}\theta$ as common in both numerator and denominator. Write $\sec \theta$ and $\operatorname{cosec}\theta$ in terms of $\sin \theta$ and $\cos \theta$ . Put the value of $\tan \theta$ and solve it further.

Complete step-by-step answer:
Let us proceed with the LHS of the given expression.
In LHS we have, $\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }$……………(1)
${{\operatorname{cosec}}^{2}}\theta$ is given in the numerator as well as the denominator.
Taking ${{\operatorname{cosec}}^{2}}\theta$ common in the numerator as well as the denominator.
Solving equation (1), we get

& \dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta } \\\ & =\dfrac{\cos e{{c}^{2}}\theta \left( 1-\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta } \right)}{\cos e{{c}^{2}}\theta \left( 1+\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta } \right)} \\\ \end{aligned}$$ $$=\dfrac{1-\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta }}{1+\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta }}$$…………(2) We know that, $$\sec \theta =\dfrac{1}{\cos \theta }$$………..(3) We also know that, $$cosec\theta =\dfrac{1}{sin\theta }$$…………………(4) Using equation (3) and equation (4), we can write equation (2) as $$\dfrac{1-\dfrac{\dfrac{1}{{{\cos }^{2}}\theta }}{\dfrac{1}{{{\sin }^{2}}\theta }}}{1+\dfrac{\dfrac{1}{{{\cos }^{2}}\theta }}{\dfrac{1}{{{\sin }^{2}}\theta }}}$$ On solving we get, $$=\dfrac{1-\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}{1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}$$…………………..(5) We know that, $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$$…………(6) Using equation (6), we can write equation (5) as $$\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$$……………(7) According to the question, the value of tan θ is given. $$\tan \theta =\dfrac{1}{\sqrt{7}}$$ Putting the value tan θ in equation (7), we get $$\begin{aligned} & \dfrac{1-{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}}{1+{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}} \\\ & =\dfrac{\dfrac{7-1}{7}}{\dfrac{7+1}{7}} \\\ & =\dfrac{6}{8} \\\ & =\dfrac{3}{4} \\\ \end{aligned}$$ So, $$\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$$. Therefore, LHS=RHS. Hence, proved. Note: This question can also be solved by putting the value of $$\operatorname{cosec}\theta $$ and $$\sec \theta $$ .But the values of $$\operatorname{cosec}\theta $$ and $$\sec \theta $$ is not given in the question. So, by using the value of $$\tan \theta $$ , we can find the value of $$\operatorname{cosec}\theta $$ and $$\sec \theta $$ .  $$tan\theta =\dfrac{1}{\sqrt{7}}$$ Using Pythagora's theorem, we can find hypotenuses. Hypotenuse= $$\sqrt{{{\left( height \right)}^{2}}+{{\left( base \right)}^{2}}}$$ $$\begin{aligned} & =\sqrt{{{\left( 1 \right)}^{2}}+{{\left( \sqrt{7} \right)}^{2}}} \\\ & =\sqrt{1+7} \\\ & =\sqrt{8} \\\ \end{aligned}$$ $$\operatorname{cosec}\theta =\dfrac{\sqrt{8}}{1}$$ and $$\sec \theta =\dfrac{\sqrt{8}}{\sqrt{7}}$$ . And after putting the value of $$\operatorname{cosec}\theta $$ and $$\sec \theta $$ in the given expression, we can get our required answer.