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# The value of ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta$ is equal to $\left( a \right){{\tan }^{2}}\theta +{{\cot }^{2}}\theta$$\left( b \right){{\sec }^{2}}\theta .{{\operatorname{cosec}}^{2}}\theta$$\left( c \right)\sec \theta .\operatorname{cosec}\theta$$\left( d \right){{\sin }^{2}}\theta .{{\cos }^{2}}\theta$$\left( e \right)1$

Last updated date: 14th Jun 2024
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Hint: We have to evaluate the value of ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta$ first and we use reciprocal identity $\sec \theta =\dfrac{1}{\cos \theta }$ and $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }.$ Then we add the term by using the identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and in the end we again use the reciprocal identity to get the solution in standard form as asked.

We have to find the value of ${{\operatorname{cosec}}^{2}}\theta +{{\sec }^{2}}\theta .$ We know that $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta },$ and we have seen that $\sec \theta =\dfrac{1}{\cos \theta }.$
Using this in ${{\operatorname{cosec}}^{2}}\theta +{{\sec }^{2}}\theta ,$ we have,
${{\operatorname{cosec}}^{2}}\theta +{{\sec }^{2}}\theta ={{\left( \dfrac{1}{\sin \theta } \right)}^{2}}+{{\left( \dfrac{1}{\cos \theta } \right)}^{2}}$
As ${{\left( \theta \right)}^{2}}={{\theta }^{2}}$ we get,
$\Rightarrow {{\operatorname{cosec}}^{2}}\theta +{{\sec }^{2}}\theta =\dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{1}{{{\cos }^{2}}\theta }$
Now we take the LCM of ${{\sin }^{2}}\theta$ and ${{\cos }^{2}}\theta$ to add the terms
$\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }$
As we know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ we get,
$\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }$
We also know that ${{a}^{2}}={{\left( a \right)}^{2}}$
$\Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}{{\left( \dfrac{1}{\cos \theta } \right)}^{2}}$
Again using the reciprocal identity, we know that,
$\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta$
$\dfrac{1}{\cos \theta }=sec\theta$
We get,
$={{\left( \operatorname{cosec}\theta \right)}^{2}}.{{\left( \sec \theta \right)}^{2}}$
$={{\operatorname{cosec}}^{2}}\theta .{{\sec }^{2}}\theta$
So, the correct answer is “Option B”.

Note: We can check how other options are not correct solutions. Let us take $\theta ={{45}^{\circ }}.$
$\left( a \right){{\tan }^{2}}\theta +{{\cot }^{2}}\theta ={{\tan }^{2}}{{45}^{\circ }}+{{\cot }^{2}}{{45}^{\circ }}$
We know that $\tan {{45}^{\circ }}=\cot {{45}^{\circ }}=1.$
$\Rightarrow 1+1=2$
While at $\theta ={{45}^{\circ }},$
${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta ={{\sec }^{2}}{{45}^{\circ }}+{{\operatorname{cosec}}^{2}}{{45}^{\circ }}$
We know that $\sec {{45}^{\circ }}=\cos {{45}^{\circ }}=\sqrt{2}$
$\Rightarrow {{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta ={{\left( \sqrt{2} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}$
$\Rightarrow {{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =2+2$
$\Rightarrow {{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =4........\left( i \right)$
Therefore, we get that ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta$ is not equal to ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta .$
$\left( c \right)\sec \theta .\operatorname{cosec}\theta$
At $\theta ={{45}^{\circ }},$ we know that,
$\Rightarrow \sec {{45}^{\circ }}=\operatorname{cosec}{{45}^{\circ }}=\sqrt{2}$
So, we get,
$\sec \theta .\operatorname{cosec}\theta =\sqrt{2}\times \sqrt{2}$
$\Rightarrow \sec \theta .\operatorname{cosec}\theta =2$
Again using (i) and the above value, we get, $\sec \theta .\operatorname{cosec}\theta$ is not equal to ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .$
$\left( d \right){{\sin }^{2}}\theta .{{\cos }^{2}}\theta$
At $\theta ={{45}^{\circ }},$ we get,
${{\sin }^{2}}\theta .{{\cos }^{2}}\theta ={{\sin }^{2}}{{45}^{\circ }}.{{\cos }^{2}}{{45}^{\circ }}$
As, $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},$ we get,
$\Rightarrow {{\sin }^{2}}\theta .{{\cos }^{2}}\theta ={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\times {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}$
$\Rightarrow {{\sin }^{2}}\theta .{{\cos }^{2}}\theta =\dfrac{1}{2}\times \dfrac{1}{2}$
Solving further, we get, ${{\sin }^{2}}\theta .{{\cos }^{2}}\theta =\dfrac{1}{4}$ at $\theta ={{45}^{\circ }}.$
So using (i) and above, we again get, ${{\sin }^{2}}\theta .{{\cos }^{2}}\theta$ is not equal to ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .$
(e) 1
At $\theta ={{45}^{\circ }},$ 1 is always 1.
But from (i), we have
${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =4$ at $\theta ={{45}^{\circ }}.$
So, 1 is not equal to ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .$
So, 1 is not equal to ${{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .$