# Prove the following; $(1{\text{ + cot A - cosec A)(1 + tan A + sec A) = 2}}$

Last updated date: 26th Mar 2023

•

Total views: 306.3k

•

Views today: 4.83k

Answer

Verified

306.3k+ views

Hint: In order to solve this question easily we will transform the given terms in of sin and cos. In this question we have to prove that the left - hand side is equal to the right - hand side.

Complete step-by-step answer:

Now, by using trigonometric identities we will easily solve the given problem. We know that $\sin {\text{A = }}\dfrac{1}{{\cos ec{\text{A}}}}$, ${\text{cosA = }}\dfrac{1}{{sec{\text{A}}}}$, ${\text{cotA = }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$, ${\text{tanA = }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$. By using these properties, we will solve this question. Now, putting these values in the given question, we get

L. H. S = $\left( {1 + {\text{ }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}{\text{ - }}\dfrac{1}{{\sin {\text{A}}}}} \right)\left( {1{\text{ + }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}{\text{ + }}\dfrac{1}{{\cos {\text{A}}}}} \right)$

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{\sin {\text{A + cosA - 1}}}}{{\sin {\text{A}}}}} \right)\left( {\dfrac{{\cos {\text{A + sinA + 1}}}}{{\cos {\text{A}}}}} \right)\]

Now we can see that in the above equation we can use the property \[{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b)(a + b)}}\]

So, applying this property we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{{{(\sin {\text{A + cosA)}}}^2}{\text{ }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

By solving further, we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{({{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A + 2cosA sinA) }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{2\sin {\text{A cosA}}}}{{\sin {\text{A cosA}}}}} \right)\] as ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$

\[ \Rightarrow \] L. H. S = 2 = R. H. S

Hence proved.

Note: To solve questions which include trigonometric terms it is suggested that you should simplify the given term by converting it into sin or cos whichever is possible. Converting in sin or cos simplify the term and you can easily solve the given term. Use identity ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$ properly after converting the trigonometric term.

Complete step-by-step answer:

Now, by using trigonometric identities we will easily solve the given problem. We know that $\sin {\text{A = }}\dfrac{1}{{\cos ec{\text{A}}}}$, ${\text{cosA = }}\dfrac{1}{{sec{\text{A}}}}$, ${\text{cotA = }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$, ${\text{tanA = }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$. By using these properties, we will solve this question. Now, putting these values in the given question, we get

L. H. S = $\left( {1 + {\text{ }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}{\text{ - }}\dfrac{1}{{\sin {\text{A}}}}} \right)\left( {1{\text{ + }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}{\text{ + }}\dfrac{1}{{\cos {\text{A}}}}} \right)$

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{\sin {\text{A + cosA - 1}}}}{{\sin {\text{A}}}}} \right)\left( {\dfrac{{\cos {\text{A + sinA + 1}}}}{{\cos {\text{A}}}}} \right)\]

Now we can see that in the above equation we can use the property \[{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b)(a + b)}}\]

So, applying this property we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{{{(\sin {\text{A + cosA)}}}^2}{\text{ }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

By solving further, we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{({{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A + 2cosA sinA) }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{2\sin {\text{A cosA}}}}{{\sin {\text{A cosA}}}}} \right)\] as ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$

\[ \Rightarrow \] L. H. S = 2 = R. H. S

Hence proved.

Note: To solve questions which include trigonometric terms it is suggested that you should simplify the given term by converting it into sin or cos whichever is possible. Converting in sin or cos simplify the term and you can easily solve the given term. Use identity ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$ properly after converting the trigonometric term.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

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

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

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

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE