 # If tanβ=(nsinαcosα)/(1-ncos²α),then tan(α+β) is equal toa.(n-1)tanαb.(n+1)tanαc.tanα/(n+1)d.-tanα/(n-1) 10 years ago

Hi Menka,

There are two ways to solve this:

Assume alpha = x, and beta = y (for the sake of typing I'm using this notation)

1.Typical Aproach:

tan y = (n sinx cosx) / (1 - n cos^2 x)

Divide Nr and Dr of RHS by cos^2 x.

======>  tan y = (n tanx)/(sec^2 x - n)

ie tan y = (n tanx) / (1+tan^2 x - n)

Now tan (x+y) = (tanx + tany) / (1- tanx tany)

Substitute for tany  ======>   tan(x+y) = [tanx + ntanx/(1+tan^2 x - n)] / [1 - tanx (n tanx/{1 + tan^2 x - n})]

Simplify =====>   tan (x+y) = tanx / (1-n)

Hence option D.

2nd Approach.

(In case questions like this appears in IIT JEE this is the appraoch you can take for a very very fast answer)

take n = 1, that would give tany = cotx = 1/tanx which means tanx*tany = 1, or tan (x+y) should have denominator as "0".

Which straight away leaves you with answer D, as only in D denominator is "0" for n = 1.

By doing this, you would have solved this question in less than a minute, leaving you with plenty of time for other questions.

And remember, IITs do give questions like this (around 5 questions in Maths), which you can finish off in less than 5 mins in all, and devote more time for other tricky questions, or could use up the time, to solve questions of other subjects.

All the Best,

Regards,