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)

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.

Hope that was helpful.

All the Best,

Regards,

Ashwin (IIT Madras).

Approved