Answer
Verified
486.6k+ views
Hint: Transform the whole equation in terms of \[\sin \theta \] and \[\cos \theta \] and then convert into desired form.
We have to prove that \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}....\left( i \right)\]
Taking \[LHS\] of equation \[\left( i \right)\], we get
\[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
Therefore, \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]
\[={{\sin }^{2}}A+{{\sec }^{2}}A+2\sin A\sec A+{{\cos }^{2}}A+{{\operatorname{cosec}}^{2}}A+2\cos A\operatorname{cosec}A\]
Rearranging the equation, we get
\[\Rightarrow \left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A....\left( ii \right)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]
Putting this value in equation \[\left( ii \right)\].
We get \[1+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A...\left( iii \right)\]
We know that \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]
We will put the values of \[\sec A\] and \[\operatorname{cosec}A\] in equation \[\left( iii \right)\].
We get, \[1+2\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A} \right)\]
\[=1+2\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right)+\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].
Hence we get,
\[1+2\left( \dfrac{1}{\sin A\cos A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]
We know, \[\dfrac{1}{\sin A}=\operatorname{cosec}A\] and \[\dfrac{1}{\cos A}=\sec A\]
Hence, we get \[1+2\operatorname{cosec}A\sec A+{{\operatorname{cosec}}^{2}}A{{\sec }^{2}}A\]
We can write it as
\[{{\left( 1 \right)}^{2}}+{{\left( \operatorname{cosec}A\sec A \right)}^{2}}+2\left( \operatorname{cosec}A\sec A \right).1\]
We know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]
By considering \[a=1\] and \[\operatorname{cosec}A\sec A=B\]
We finally get \[LHS={{\left( 1+\operatorname{cosec}A\sec A \right)}^{2}}\] which is equal to \[\text{RHS}\].
Hence Proved.
Note: By looking the terms of \[\operatorname{cosec}A\] and \[secA\] in \[\text{RHS}\], students convert \[\sin A\] and \[\cos A\] into \[\dfrac{1}{\operatorname{cosec}A}\] and \[\dfrac{1}{secA}\] respectively in first step only, but that creates confusion and does not give the desired results.
We have to prove that \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}....\left( i \right)\]
Taking \[LHS\] of equation \[\left( i \right)\], we get
\[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
Therefore, \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]
\[={{\sin }^{2}}A+{{\sec }^{2}}A+2\sin A\sec A+{{\cos }^{2}}A+{{\operatorname{cosec}}^{2}}A+2\cos A\operatorname{cosec}A\]
Rearranging the equation, we get
\[\Rightarrow \left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A....\left( ii \right)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]
Putting this value in equation \[\left( ii \right)\].
We get \[1+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A...\left( iii \right)\]
We know that \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]
We will put the values of \[\sec A\] and \[\operatorname{cosec}A\] in equation \[\left( iii \right)\].
We get, \[1+2\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A} \right)\]
\[=1+2\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right)+\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].
Hence we get,
\[1+2\left( \dfrac{1}{\sin A\cos A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]
We know, \[\dfrac{1}{\sin A}=\operatorname{cosec}A\] and \[\dfrac{1}{\cos A}=\sec A\]
Hence, we get \[1+2\operatorname{cosec}A\sec A+{{\operatorname{cosec}}^{2}}A{{\sec }^{2}}A\]
We can write it as
\[{{\left( 1 \right)}^{2}}+{{\left( \operatorname{cosec}A\sec A \right)}^{2}}+2\left( \operatorname{cosec}A\sec A \right).1\]
We know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]
By considering \[a=1\] and \[\operatorname{cosec}A\sec A=B\]
We finally get \[LHS={{\left( 1+\operatorname{cosec}A\sec A \right)}^{2}}\] which is equal to \[\text{RHS}\].
Hence Proved.
Note: By looking the terms of \[\operatorname{cosec}A\] and \[secA\] in \[\text{RHS}\], students convert \[\sin A\] and \[\cos A\] into \[\dfrac{1}{\operatorname{cosec}A}\] and \[\dfrac{1}{secA}\] respectively in first step only, but that creates confusion and does not give the desired results.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE