Question

If $\sin A+\cos A=\sqrt{2}\sin \left( 90-A \right)$, then obtain the value of $\cot A$.

Answer 1

Hint: To solve this question, we will start from the given equality. We should know that $\sin \left( 90-A \right)=\cos A$ and $\dfrac{\sin A}{\cos A}=\tan A$. We should also know that $\dfrac{1}{\tan A}=\cot A$. By using these properties, we will be able to find the value of $\cot A$.

Complete step-by-step answer:

In this question, we have been asked to find the value of $\cot A$. And it is given in the question that $\sin A+\cos A=\sqrt{2}\sin \left( 90-A \right)$. So, to obtain the value of $\cot A$, we will start with the given equality, which is, $\sin A+\cos A=\sqrt{2}\sin \left( 90-A \right)$. We know that $\sin \left( 90-A \right)=\cos A$, so we can apply that in the above equality and write the given equality as,
$\sin A+\cos A=\sqrt{2}\cos A$
Now, we will take the like terms with $\cos A$ to the right hand side. So, we can write the above equation as, $\sin A=\sqrt{2}\cos A-\cos A$
Now, we will take $\cos A$ as common. By taking $\cos A$as common, we can write the above equation as, $\sin A=\left( \sqrt{2}-1 \right)\cos A$
We will now divide the whole equation by $\cos A$. So, we can write the above equation as, $\dfrac{\sin A}{\cos A}=\left( \sqrt{2}-1 \right)\dfrac{\cos A}{\cos A}$
By cancelling $\cos A$ in the numerator as well as denominator as it is common, we get, $\dfrac{\sin A}{\sin B}=\sqrt{2}-1$
We know that, $\dfrac{\sin A}{\cos A}=\tan A$, so by applying that in the above equation, we get, $\tan A=\sqrt{2}-1$. Now, we also know that, $\dfrac{1}{\tan A}=\cot A$, so by applying that in the above equation, we get, $\dfrac{1}{\cot A}=\sqrt{2}-1$.
Now we will reciprocate the whole equation. By reciprocating the above equation, we get, $\cot A=\dfrac{1}{\sqrt{2}-1}$
Now, we will rationalize $\dfrac{1}{\sqrt{2}-1}$ by multiplying the numerator and denominator by $\sqrt{2}+1$ and get,
$\begin{align}
  & \cot A=\dfrac{1}{\left( \sqrt{2}-1 \right)}\times \dfrac{\left( \sqrt{2}+1 \right)}{\left( \sqrt{2}+1 \right)} \\
 & \Rightarrow \cot A=\dfrac{\sqrt{2}+1}{2-1} \\
 & \Rightarrow \cot A=\sqrt{2}+1 \\
\end{align}$
Hence, we have obtained the value of $\cot A$ as $\sqrt{2}+1$.

Note: We should keep in mind that whenever we are left with irrational numbers in the denominator, then we have to rationalize them. For example, if we get $a-\sqrt{b}$ in the denominator, then we should multiply the numerator and denominator by $a+\sqrt{b}$.