Question

Prove that $1 + cotAtanB = 0$ if we have given $sinAsinB - cosAcosB + 1 = 0$ ?

Answer 1

We will try to simplify the equation $sinAsinB - cosAcosB + 1 = 0$ and try to find the range and value of a and b. then we will simplify the equation $1 + cotAtanB = 0$ and if this equation satisfies the value of a and b then it stands for true. But most important thing for their simplification is use of trigonometric equation.

Complete step by step answer:
We have given that $sinAsinB - cosAcosB + 1 = 0$
We will simplify the above equation and try to find the value of a and b
$\Rightarrow sinAsinB - cosAcosB + 1 = 0$
We just simplify them
$\Rightarrow cosAcosB - sinAsinB = 1$
We know that $cosAcosB - sinAsinB = \cos (A + B)$
$\Rightarrow \cos (A + B) = 1$
We know that $\cos 0 = 1$
$\Rightarrow (A + B) = 0$ ---(1)
Now, we will simplify the equation $1 + cotAtanB = 0$ and try to prove it
$\Rightarrow 1 + cotAtanB = 0$
We know that cot is the ratio of cos and sin while tan is the ratio of sine and cos.
$\Rightarrow 1 + \dfrac{{\cos A}}{{\sin A}} \times \dfrac{{\sin B}}{{\cos B}} = 0$
$\Rightarrow \dfrac{{\sin A\cos B + \cos A\sin B}}{{\sin A\cos B}} = 0$
We know that$\sin a\cos b + \cos a\sin b = \sin (a + b)$ $\sin a\cos b + \cos a\sin b = \sin (a + b)$
$\Rightarrow \dfrac{{\sin (A + B)}}{{\sin A\cos B}} = 0$
$\Rightarrow \sin (A + B) = 0$ -- (2)
We have already found the value of $(A + B) = 0$,
We know that $\sin 0 = 0$,
$\Rightarrow \sin (A + B) = 0$
So, equation 2 is true.
Hence, we have proved that $1 + cotAtanB = 0$ if $sinAsinB - cosAcosB + 1 = 0$ .

Note:
We have to be familiar with different trigonometric equations for solving these types of questions. We should be familiar with equations like $cosAcosB - sinAsinB = \cos (A + B)$ , $\sin a\cos b + \cos a\sin b = \sin (a + b)$ etc.
We should note that this type of problem can be solved by taking the proving expression first and reaching out to the given expression.