Question

Using the trigonometric identities, prove that:
$ \dfrac{{\cot A + \tan B}}{{\cot B + \tan A}} = \cot A.\tan B $

Answer 1

Hint: To solve the given problem, take the left-hand side of the given expression and simplify it by converting it into sine and cosine. Then, take the L.C.M. in numerator and denominator.

Complete step-by-step solution -
Take, the left-hand side of the given expression and simplify as shown below.
$\dfrac{{\cot A + \tan B}}{{\cot B + \tan A}} \\$
 $\Rightarrow \dfrac{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin B}}{{\cos B}}}}{{\dfrac{{\cos B}}{{\sin B}} + \dfrac{{\sin A}}{{\cos A}}}} \\ $
Now, take the L.C.M. in numerator and denominator of the above expression.
$ \dfrac{{\dfrac{{\cos A\cos B + \sin A\sin B}}{{\sin A\cos B}}}}{{\dfrac{{\cos B\cos A + \sin A\sin B}}{{\sin B\cos A}}}} \\$
$\Rightarrow \dfrac{{\sin B\cos A}}{{\cos B\sin A}} \\$
$\Rightarrow \left( {\dfrac{{\sin B}}{{\cos B}}} \right)\left( {\dfrac{{\cos A}}{{\sin A}}} \right) \\$
$\Rightarrow \tan B\cot A \\ $
Hence proved.

Note: In the given problem, convert the trigonometric identities into sine and cosine. Here we can use a simpler approach to solve this problem as we can convert $\cot A$ and $\cot B$ respectively in $\tan A$ and $\tan B$ and cancel the like terms from numerator and denominator.