Answer
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Hint: In this question we have to prove that LHS=RHS. First we will consider the LHS of the given expression and by simplifying prove it equal to RHS. We will use the trigonometric ratios to simplify the left hand side of the equation.
Complete step by step solution:
We have been given an expression \[\left( \sin A+\cos A \right)\left( \sec A+cosecA \right)=2+\sec A.cosecA\].
We have to prove that the left hand side is equal to right hand side.
Let us first consider the LHS, then we will get
\[\Rightarrow \left( \sin A+\cos A \right)\left( \sec A+cosecA \right)\]
Now, opening the parenthesis we will get
\[\Rightarrow \sin A.\sec A+\cos A.\sec A+\sin A.cosecA+\cos A.cosecA\]
Now, we know that \[cosecA=\dfrac{1}{\sin A}\] and $\sec A=\dfrac{1}{\cos A}$ .
Now, substituting the values in the above obtained equation we will get
\[\Rightarrow \sin A.\dfrac{1}{\cos A}+\cos A.\dfrac{1}{\cos A}+\sin A.\dfrac{1}{\sin A}+\cos A.\dfrac{1}{\sin A}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
& \Rightarrow \dfrac{\sin A}{\cos A}+1+1+\dfrac{\cos A}{\sin A} \\
& \Rightarrow \dfrac{{{\sin }^{2}}A+\sin A.\cos A+\sin A.\cos A+{{\cos }^{2}}A}{\sin A.\cos A} \\
\end{align}\]
Now, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Now, substituting the value in the above obtained equation we will get
\[\Rightarrow \dfrac{1+2\sin A.\cos A}{\sin A.\cos A}\]
We can rewrite the above equation as
\[\Rightarrow \dfrac{1}{\sin A.\cos A}+2\]
Now, substituting the values we will get
$\Rightarrow 2+\sec A.\operatorname{cosec}A$
So we get LHS=RHS
Hence proved
Note: Alternatively we can consider the right hand side of the expression and simplify it to prove it equal to the left hand side. But sometimes it becomes lengthy so it is advisable that we have to consider the LHS and simplify it. To solve this type of questions students must have knowledge of trigonometric formulas, ratios and identities. Students must know the relation between different trigonometric functions to convert one into another form.
Complete step by step solution:
We have been given an expression \[\left( \sin A+\cos A \right)\left( \sec A+cosecA \right)=2+\sec A.cosecA\].
We have to prove that the left hand side is equal to right hand side.
Let us first consider the LHS, then we will get
\[\Rightarrow \left( \sin A+\cos A \right)\left( \sec A+cosecA \right)\]
Now, opening the parenthesis we will get
\[\Rightarrow \sin A.\sec A+\cos A.\sec A+\sin A.cosecA+\cos A.cosecA\]
Now, we know that \[cosecA=\dfrac{1}{\sin A}\] and $\sec A=\dfrac{1}{\cos A}$ .
Now, substituting the values in the above obtained equation we will get
\[\Rightarrow \sin A.\dfrac{1}{\cos A}+\cos A.\dfrac{1}{\cos A}+\sin A.\dfrac{1}{\sin A}+\cos A.\dfrac{1}{\sin A}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
& \Rightarrow \dfrac{\sin A}{\cos A}+1+1+\dfrac{\cos A}{\sin A} \\
& \Rightarrow \dfrac{{{\sin }^{2}}A+\sin A.\cos A+\sin A.\cos A+{{\cos }^{2}}A}{\sin A.\cos A} \\
\end{align}\]
Now, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Now, substituting the value in the above obtained equation we will get
\[\Rightarrow \dfrac{1+2\sin A.\cos A}{\sin A.\cos A}\]
We can rewrite the above equation as
\[\Rightarrow \dfrac{1}{\sin A.\cos A}+2\]
Now, substituting the values we will get
$\Rightarrow 2+\sec A.\operatorname{cosec}A$
So we get LHS=RHS
Hence proved
Note: Alternatively we can consider the right hand side of the expression and simplify it to prove it equal to the left hand side. But sometimes it becomes lengthy so it is advisable that we have to consider the LHS and simplify it. To solve this type of questions students must have knowledge of trigonometric formulas, ratios and identities. Students must know the relation between different trigonometric functions to convert one into another form.
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