Question

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

Answer 1

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.

Complete step-by-step answer:
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 \[{{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .\]