Answer

Verified

412.2k+ views

**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 \[{{\sec }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta .\]

Recently Updated Pages

What are the Advantages and Disadvantages of Algorithm

How do you write 0125 in scientific notation class 0 maths CBSE

The marks obtained by 50 students of class 10 out of class 11 maths CBSE

Out of 30 students in a class 6 like football 12 like class 7 maths CBSE

Explain the law of constant proportion in a simple way

How do you simplify left 5 3i right2 class 12 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Mention the different categories of ministers in the class 10 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Who is the executive head of the Municipal Corporation class 6 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which monarch called himself as the second Alexander class 10 social science CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE