1I got it
u multiply a^3 with b^3 that will giv sinθ.cosθ..
now u add a^6+b^6.
at the end you will have sinθ.cosθ at denominator which terminates as a^3.b^3
Wanted to know whats the main point to keep in mind when solving trigonometries.
Most of the time i find it REALLY difficult. for e.g. i jus could not solve this
eliminate θ from the following
sinθ.sinθ/cosθ= a^3 and cosθ.cosθ/sinθ=b^3
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3 Answers
62Tough one..
As a beginner i just din like this chapter.. found it very uninspiring which i first read it.. I was thrown 1000 formulas to mug up .. made a chart.. put it on my study desk and never read it again :P
So what you need to do is to see small small things..
There are lots of patterns/ clues within the question..
For instance one clue in this question is that if we are able to find sin and cos seperately we can sq;uare adn add them to zero.. (may not be the best clue..)
1
1By multiplying a3 and b3 we get a3b3=sinθcosθ.
Then,
(a3)2 + (b3)2
= sin4θ /cos2θ + cos4θ/sin2θ
= [sin^6(θ) + cos^6(θ)] / sin2θ.cos2θ
= [(sin2θ)3 + (cos2θ)3] / sin2θ.cos2θ
= (sin2θ+cos2θ)(sin4θ+cos4θ - sin2θ.cos2θ) / sin2θ.cos2θ
= [sin4θ + cos4θ - sin2θ.cos2θ ] / sin2θ.cos2θ
= [1 - 3sin2θ.cos2θ]/sin2θ.cos2θ
Substitute the value a3b3=sinθcosθ in the above equation and u will get the answer...
(Excuse any mistakes!)