Question

Prove that \[1 + cotAtanB = 0\] if we have given \[sinAsinB - cosAcosB + 1 = 0\] ?

Answer 1

Hint: 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.