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If tanθtanα=√[(a-b)/(a+b)] ,then the value of (a-bcos2α)(a-bcos2θ) is a.(a²-b²)tanα b.(a²-b²)tanθ c.a²-b² d.a²+b²

If tanθtanα=√[(a-b)/(a+b)],then the value of (a-bcos2α)(a-bcos2θ) is


a.(a²-b²)tanα


b.(a²-b²)tanθ


c.a²-b²


d.a²+b²

Grade:12th Pass

1 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Menka,

 

Here again, lets look at two solutions to this problem.

One is the conventional approach, and the other how to solve this problem if asked in IIT JEE.

 

For sake of easy typing, let alpha = x, and theta = y.

 

1. Conventional approach (would be lengthy, so suggest you to solve with a paper/pen side-by-side)

 

Now [tanx tany]^2 = (a-b)/(a+b)

So (a+b) tan^2 x = (a-b) cot^2 y

ie a(cot^2 y - tan^2x) = b(cot^2 y + tan^2 x)

ie b/a = (cot^2 y - tan^2 x)/(cot^2 y + tan^2 x) ---------------- (1)

 

Next consider the expression in the question and take "a" common from both the brackets, and you will have

a^2 [1 - b/a cos 2x]*[1 - b/a cos 2y] ------------ (2)

 

Now, cos 2x = (1 - tan^2 x)/(1 + tan^2 x)

So from (1) b/a cos2x = Substitute for b/a and cos 2x, and multiply, you will get

 

[1 - b/a cos 2x] = 2(1+tan^2 y) / [(cot^2 y + tan^2 x)*(1+ tan^2 x)

 

Similarly [1 - b/a cos 2y] = 2[(tan^2 y)*(1+cot^2 x)] / [(cot^2 y + tan^2 x)*(1 + tan^2 y)].

 

So expression in (2) would reduce to [ (4a^2) (tan^2 y) (1 + cot^2 x) ] / [ (1+tan^2 x) (cot^2 y + tan^2x)^2 ].

which is nothing but [ (4a^2) (tan^2 x)(tan^2 y) ] / [ 1 + (tann^2 x)(tan^2 y) ]

 

But (tan^2 x)(tan^2 y) = (a-b)/(a+b)

Substitute in the above expression, and you will easily get, (a^2 - b^2).

[You should solve with a paper and pen, to get the idea more clearly]

 

 

Now the above method would be the actual solving method.

But if you solve like this you will not take less than 10 mins to solve this question (or may be even end up spending 5 mins, and not solving it and hence leave it un-attempted).....

For such questions, we use a shortcut, which would give you the answer in less than 2 minutes.

Let's see how to go about that method, if you get such questions in IIT JEE.

 

 

Now check what happens when alpha = theta, ie when x = y,

You will have tan^2 x = root [(a-b)/(a+b)]

And the expression is [a - b cos 2x]^2

where cos2x = (1 - tan^2 x) / (1 + tan^2 x) = [ root(a+b) - root(a-b) ] / [ root(a+b) + root(a-b) ]

which is nothing but [ a - root(a^2 - b^2) ] / b --------- (by multiplying the Dr with its conjugate)

So b*cos2x = a - root(a^2 - b^2), which will =====>  a - b*cos2x = root(a^2 - b^2).

And Hence the answer is the square of that, which is a^2 - b^2.

 

Hope that helps.

 

All the best,

Regards,

Ashwin (IIT Madras).

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