Question

If tan B= nsin A cos A÷1-n cos^2 A
then tan(A + B) equals
​

Answer 1

Answer:

$tan(A+B)=\frac{tanA}{1-n}$

Step-by-step explanation:

Formula used:

$tan(A+B)=\frac{tanA+tanB}{1-tanA\:tanB}$

Given:

$tanB=\frac{n\:sinA\:cosA}{1-n\:cos^2A}$

Divde both Nr. and Dr. by $cos^2A$

$tanB=\frac{\frac{n\:sinA\:cosA}{cos^2A}}{\frac{1-n\:cos^2A}{cos^2A}}$

$tanB=\frac{\frac{n\:sinA}{cosA}}{sec^2A-n}$

$tanB=\frac{n\:tanA}{1+tan^2A-n}$

Now,

$tan(A+B)=\frac{tanA+tanB}{1-tanA\:tanB}$

$tan(A+B)=\frac{tanA+\frac{n\:tanA}{1+tan^2A-n}}{1-tanA\:\frac{n\:tanA}{1+tan^2A-n}}$

$tan(A+B)=\frac{\frac{tanA+tan^3A-n\:tanA+n\:tanA}{1+tan^2A-n}}{\frac{1+tan^2A-n-n\:tan^2A}{1+tan^2A-n}}$

$tan(A+B)=\frac{tanA+tan^3A-n\:tanA+n\:tanA}{1+tan^2A-n-n\:tan^2A}$

$tan(A+B)=\frac{tanA+tan^3A}{(1+tan^2A)-n(1+tan^2A)}$

$tan(A+B)=\frac{tanA+tan^3A}{(1+tan^2A)(1-n)}$

$tan(A+B)=\frac{tanA(1+tan^2A)}{(1+tan^2A)(1-n)}$

$tan(A+B)=\frac{tanA}{1-n}$