Question

Find the value of the following expression using the trigonometric identities and properties
$\dfrac{{{\sec }^{2}}\left( 90-\theta \right)-{{\cot }^{2}}\theta }{2\left( {{\sin }^{2}}{{25}^{{}^\circ }}+{{\sin }^{2}}{{65}^{{}^\circ }} \right)}+\dfrac{2{{\sin }^{2}}{{30}^{{}^\circ }}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$.

Answer 1

Hint:We will evaluate the given trigonometric function using some basic trigonometry formulas and trigonometric standard angles. Following are some of the formula used
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$
${{\cot }^{2}}\theta +1={{\operatorname{cosec}}^{2}}\theta$, also
Some trigonometric conversions are as follows
$\sin \left( 90-\theta \right)=\cos \theta$
$\cos \left( 90-\theta \right)=\sin \theta$
$\tan \left( 90-\theta \right)=\cot \theta$
$\cot \left( 90-\theta \right)=\tan \theta$
$\sec \left( 90-\theta \right)=\operatorname{cosec}\theta$
$\operatorname{cosec}\left( 90-\theta \right)=\sec \theta$
Apply these formulas and simplify it to get the value for the question.

Complete step-by-step answer:
It is given in the question that we have to evaluate the following expression $\dfrac{{{\sec }^{2}}\left( 90-\theta \right)-{{\cot }^{2}}\theta }{2\left( {{\sin }^{2}}{{25}^{{}^\circ }}+{{\sin }^{2}}{{65}^{{}^\circ }} \right)}+\dfrac{2{{\sin }^{2}}{{30}^{{}^\circ }}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$.
We know that $\sec \left( 90-\theta \right)=\operatorname{cosec}\theta$ and $\sin \left( 90-\theta \right)=\cos \theta$. So, we can write ${{\sec }^{2}}\left( 90-\theta \right)={{\operatorname{cosec}}^{2}}\theta$ and ${{\sin }^{2}}{{65}^{{}^\circ }}={{\sin }^{2}}\left( {{90}^{{}^\circ }}-{{65}^{{}^\circ }} \right)=\cos {{25}^{{}^\circ }}$ we get
= $\dfrac{{{\operatorname{cosec}}^{2}}\left( \theta \right)-{{\cot }^{2}}\theta }{2\left( {{\sin }^{2}}{{25}^{{}^\circ }}+{{\cos }^{2}}{{25}^{{}^\circ }} \right)}+\dfrac{2{{\sin }^{2}}{{30}^{{}^\circ }}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$.
Now, we know that ${{\cot }^{2}}\theta +1={{\operatorname{cosec}}^{2}}\theta$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So, we can write ${{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1$, therefore we get $\dfrac{1}{2\left( 1 \right)}+\dfrac{2{{\sin }^{2}}{{30}^{{}^\circ }}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$. Now, we have the value of $\sin {{30}^{{}^\circ }}=\dfrac{1}{2}$, we put the value in the equation as $\dfrac{1}{2}+\dfrac{2{{\left( \dfrac{1}{2} \right)}^{2}}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$ .
We know that $\tan \left( 90-\theta \right)=\cot \theta$ and $\cot \left( 90-\theta \right)=\tan \theta$. Therefore, we can use these formulas in the second term as follow $co{{t}^{2}}{{57}^{{}^\circ }}={{\cot }^{2}}\left( {{90}^{{}^\circ }}-{{33}^{{}^\circ }} \right)={{\tan }^{2}}{{33}^{{}^\circ }}$ and ${{\tan }^{2}}{{58}^{{}^\circ }}={{\tan }^{2}}\left( {{90}^{{}^\circ }}-{{32}^{{}^\circ }} \right)={{\cot }^{2}}{{32}^{{}^\circ }}$. Putting these values in the equation $\dfrac{1}{2}+\dfrac{2{{\left( \dfrac{1}{2} \right)}^{2}}{{\tan }^{2}}{{32}^{{}^\circ }}{{\tan }^{2}}{{58}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\cot }^{2}}{{57}^{{}^\circ }} \right)}$ we get equation modified as $\dfrac{1}{2}+\dfrac{2\left( \dfrac{1}{4} \right){{\tan }^{2}}{{32}^{{}^\circ }}{{\cot }^{2}}{{32}^{{}^\circ }}}{3\left( {{\sec }^{2}}{{33}^{{}^\circ }}-{{\tan }^{2}}{{33}^{{}^\circ }} \right)}$.
Now we will use the formulas $\tan \theta \cot \theta =1$ and ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$ or ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ we get $\dfrac{1}{2}+\dfrac{2\left( \dfrac{1}{4} \right)\left( 1 \right)}{3\left( 1 \right)}=\dfrac{1}{2}+\dfrac{1}{2\times 3}$, finally solving this equation further we get
= $\dfrac{1}{2}+\dfrac{1}{6}$
= $\dfrac{3+1}{6}$
= $\dfrac{4}{6}=\dfrac{2}{3}$.
Therefore the given expression has the value of $\dfrac{2}{3}$ obtained by using basic trigonometry formulas.

Note: Student may take the value of ${{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =-1$ in hurry but this is a wrong and the correct value is ${{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1$. Usually this error arises because students are in a hurry and make the wrong transpose of trigonometry identities from LHS to RHS. Therefore it is necessary to memorize such trigonometry identities ,formulas,trigonometric conversions and trigonometric standard angles to solve these types of questions.