Question

Prove that $\dfrac{\cos A}{\left( 1+\sin A \right)}+\dfrac{1+\sin A}{\left( \cos A \right)}=2\sec A$.

Answer 1

Consider the L.H.S and take the L.C.M of the denominator of the two terms and simplify. Use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and expand the terms in the numerator. Now, use the algebraic identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to simplify the numerator further and take the common terms together. Cancel the common terms from the numerator and the denominator to get the R.H.S. Use the conversion $\dfrac{1}{\cos x}=\sec x$.

Complete step by step solution:
Here we have been provided with the L.H.S expression $\dfrac{\cos A}{\left( 1+\sin A \right)}+\dfrac{1+\sin A}{\left( \cos A \right)}$ and we are asked it to prove it equal to the R.H.S expression $2\sec A$.
Now, here we will simplify the L.H.S using some basic trigonometric identities and try to get the R.H.S, so we have,
$\Rightarrow L.H.S=\dfrac{\cos A}{\left( 1+\sin A \right)}+\dfrac{1+\sin A}{\left( \cos A \right)}$
Taking the L.C.M we get,
$\Rightarrow L.H.S=\dfrac{\left( {{\cos }^{2}}A \right)+{{\left( 1+\sin A \right)}^{2}}}{\left( 1+\sin A \right)\left( \cos A \right)}$
Using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand the term containing the sine function in the numerator we get,
$\begin{aligned} & \Rightarrow L.H.S=\dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{\left( 1+\sin A \right)\left( \cos A \right)} \\\ & \Rightarrow L.H.S=\dfrac{\left( {{\cos }^{2}}A+{{\sin }^{2}}A \right)+1+2\sin A}{\left( 1+\sin A \right)\left( \cos A \right)} \\\ \end{aligned}$
Using the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we get,
$\begin{aligned} & \Rightarrow L.H.S=\dfrac{1+1+2\sin A}{\left( 1+\sin A \right)\left( \cos A \right)} \\\ & \Rightarrow L.H.S=\dfrac{2+2\sin A}{\left( 1+\sin A \right)\left( \cos A \right)} \\\ & \Rightarrow L.H.S=\dfrac{2\left( 1+\sin A \right)}{\left( 1+\sin A \right)\left( \cos A \right)} \\\ \end{aligned}$
Cancelling the common terms from the numerator and the denominator we get,
$\Rightarrow L.H.S=\dfrac{2}{\left( \cos A \right)}$
Using the conversion $\dfrac{1}{\cos x}=\sec x$ we get,
$\begin{aligned} & \Rightarrow L.H.S=2\sec A \\\ & \therefore L.H.S=R.H.S \\\ \end{aligned}$

Note: Remember all the trigonometric identities as they will help in solving the Question in less time. There are some more basic trigonometric identities like ${{\csc }^{2}}x-{{\cot }^{2}}x=1$ and ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. You may note that here we cannot solve the R.H.S because there we have only one term which cannot be simplified easily in such a way that we might get the R.H.S. That is why we focused on the L.H.S only. In higher mathematics we have many more trigonometric identities which must be remembered.