Answer

Verified

451.5k+ views

Hint: -Here, we have the value of m and n so, we go through by simply putting the value of $m$ and $n$ to proceed further.

Given,$\;m = \left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)$ $n = {\text{tan}}\theta - {\text{sin}}\theta $

We need to show ${m^2} - {n^2} = 4\sqrt {mn} $

Taking${\text{L}}{\text{.H}}{\text{.S}}$.

${m^2} - {n^2} = {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)^2} - {\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)^2}$

Here, we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

By applying these formula,

\[

= {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta - {\tan ^2}\theta - {\sin ^2}\theta + 2\tan \theta \sin \theta \\

= 4{\text{tan}}\theta \cdot {\text{sin}}\theta \\

\]

Now, taking \[{\text{R}}{\text{.H}}{\text{.S}}\].

\[4\sqrt {mn} = 4\sqrt {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)} \]

And here we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

\[ = 4\sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \] $\left( {\because {{\tan }^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)$

\[

= 4\sqrt {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} - {{\sin }^2}\theta } \\

= 4\sqrt {\frac{{{{\sin }^2}\theta (1 - {{\cos }^2}\theta )}}{{{{\cos }^2}\theta }}} \\

\]

\[ = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \cdot {\text{si}}{{\text{n}}^2}\theta } \] \[\left( {\because (1 - {{\cos }^2}\theta ) = {{\sin }^2}\theta } \right)\]

\[ = 4\tan \theta \cdot \sin \theta \]

Therefore, \[{\text{L}}{\text{.H}}{\text{.S}}{\text{ = R}}{\text{.H}}{\text{.S}}{\text{.}}\]

Hence proved.

Note:-Whenever we face such type of question try it solving by taking ${\text{L}}{\text{.H}}{\text{.S}}$and \[{\text{R}}{\text{.H}}{\text{.S}}\] and then equate it to proof the question.

Given,$\;m = \left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)$ $n = {\text{tan}}\theta - {\text{sin}}\theta $

We need to show ${m^2} - {n^2} = 4\sqrt {mn} $

Taking${\text{L}}{\text{.H}}{\text{.S}}$.

${m^2} - {n^2} = {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)^2} - {\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)^2}$

Here, we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

By applying these formula,

\[

= {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta - {\tan ^2}\theta - {\sin ^2}\theta + 2\tan \theta \sin \theta \\

= 4{\text{tan}}\theta \cdot {\text{sin}}\theta \\

\]

Now, taking \[{\text{R}}{\text{.H}}{\text{.S}}\].

\[4\sqrt {mn} = 4\sqrt {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)} \]

And here we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

\[ = 4\sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \] $\left( {\because {{\tan }^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)$

\[

= 4\sqrt {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} - {{\sin }^2}\theta } \\

= 4\sqrt {\frac{{{{\sin }^2}\theta (1 - {{\cos }^2}\theta )}}{{{{\cos }^2}\theta }}} \\

\]

\[ = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \cdot {\text{si}}{{\text{n}}^2}\theta } \] \[\left( {\because (1 - {{\cos }^2}\theta ) = {{\sin }^2}\theta } \right)\]

\[ = 4\tan \theta \cdot \sin \theta \]

Therefore, \[{\text{L}}{\text{.H}}{\text{.S}}{\text{ = R}}{\text{.H}}{\text{.S}}{\text{.}}\]

Hence proved.

Note:-Whenever we face such type of question try it solving by taking ${\text{L}}{\text{.H}}{\text{.S}}$and \[{\text{R}}{\text{.H}}{\text{.S}}\] and then equate it to proof the question.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

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

How many crores make 10 million class 7 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE