Question

Prove that $\sin A(1+\tan A)+\cos A(1+\cot A)=\sec A+\csc A$

Answer 1

Hint: Here, first we should consider the LHS and convert tan A and cot A as:

$\tan A=\dfrac{\sin A}{\cos A}$

$\cot A=\dfrac{\cos A}{\sin A}$

Then, take outside the common factors and do the simplification to make it much simpler.

Here, we also have to apply the trigonometric identity, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$

Here, we have to prove $\sin A(1+\tan A)+\cos A(1+\cot A)=\sec A+\csc A$.

Here, first consider the LHS, $\sin A(1+\tan A)+\cos A(1+\cot A)=\sec A+\csc A$.

Complete step-by-step answer:

We know that:

$\tan A=\dfrac{\sin A}{\cos A}$

$\cot A=\dfrac{\cos A}{\sin A}$

Now, by substituting these values in the LHS, we obtain the equation:

$\sin A(1+\tan A)+\cos A(1+\cot A)=\sin A\left( 1+\dfrac{\sin A}{\cos A} \right)+\cos A\left( 1+\dfrac{\cos A}{\sin A} \right)$

Now, from the above equation we can take outside $\dfrac{1}{\cos A}$ from the first term and $\dfrac{1}{\sin A}$ from the second term. Hence, we will get the equation:

$\sin A(1+\tan A)+\cos A(1+\cot A)=\dfrac{\sin A}{\cos A}\left( \cos A+\sin A \right)+\dfrac{\cos A}{\sin A}\left( \sin A+\cos A \right)$

Here, $\sin A+\cos A$ is a common factor for both the terms. Therefore, we can take it outside

$\Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\left( \cos A+\sin A \right)\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)$

Next, by taking the LCM we get:

$\Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\left( \cos A+\sin A \right)\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\sin A} \right)$

We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

$\begin{align}

  & \Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\left( \cos A+\sin A \right)\left( \dfrac{1}{\cos A\sin A} \right) \\

 & \Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\dfrac{\cos A+\sin A}{\cos A\sin A} \\

\end{align}$

Now, by splitting the terms,

$\Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\dfrac{\cos A}{\cos A\sin A}+\dfrac{\sin

 A}{\cos A\sin A}$

Next, we have to do the cancellation.

$\Rightarrow \sin A(1+\tan A)+\cos A(1+\cot A)=\dfrac{1}{\sin A}+\dfrac{1}{\cos A}$

We know that the reciprocal of sin A is cosec A and reciprocal of cos A is sec A. That is,

$\begin{align}

  & \dfrac{1}{\sin A}=\csc A \\

 & \dfrac{1}{\cos A}=\sec A \\

\end{align}$

Now, by substituting these values in our equation, we get:

$\sin A(1+\tan A)+\cos A(1+\cot A)=\csc A+\sec A$

Hence, proved.

Note: Here, you should be familiar about the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. Without knowing this you can’t go further. There are six trigonometric ratios and from sin, cos and tan we can define other trigonometric ratios where:

$\begin{align}

  & \dfrac{1}{\sin A}=\csc A \\

 & \dfrac{1}{\cos A}=\sec A \\

 & \dfrac{1}{\tan A}=\cot A \\

\end{align}$