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Prove That [(1/sec20 - cos20) + (1/cosec20 - sin20)](sin20cos20) = 1 - 2sin20cos20/2 + sin20cos20

Prove That [(1/sec20 - cos20) + (1/cosec20 - sin20)](sin20cos20) = 1 - 2sin20cos20/2 + sin20cos20

Grade:10

1 Answers

Soumendu Majumdar
159 Points
3 years ago
Dear Student,
The way you have written the question sin^2 theta has become sin 20 as well as the other trigonometric ratios...please check the question before posting it whether it is understandable or not, Now let's move on to the solution.
We need to prove
[ { 1/( sec^2 theta – cos^2 theta) + 1/( cosec^2 theta – sin^2 theta) }(sin^2 theta cos^2 theta) ] =( 1 – sin^2 theta cos^2 theta )
divided by ( 2 + sin^2 theta cos^2 theta )
To make this short i will denote theta by A ;
so L.H.S = [ { ( sin^2 A cos^2 A)/( sec^2 A – cos^2 A) } + { (sin^2 A cos^2 A)/( cosec^2 A – sin^2 A) } ]
= [ { ( sin^2 A cos^4 A )/( 1 – cos^4 A) } + { ( sin^4 A cos^2 A )/( 1 – sin^4 A) } ]
Now break ( 1 – cos^4 A ) into ( 1 + cos^2 A )( 1 – cos^2 A) and ( 1 – cos^2 A)= sin^2 A so it can be terminated from numerator and denominator, do the same for the one with ( 1 – sin^4 A) in denominator
so now we get 
L.H.S= { (cos^4 A)/( 1 + cos^2 A ) + (sin^4 A)/( 1 + sin^2 A ) }
= { ( cos^4 A + sin^4 A + sin^2 A cos^4 A + sin^4 A cos^2 A)/( 1 + sin^2 A + cos^2 A + sin^2 A cos^2 A ) }
Now sin^4 A + cos^4 A = 1 – 2sin^2 A cos^2 A
so L.H.S= [ { 1 – 2sin^2 A cos^2 A + sin^2 A cos^2 A( sin^2 A + cos^2 A) }/{ 1 + sin^2 A + cos^2 A + sin^2 A cos^2 A } ]
Using the formula sin^2 A + cos^2 A = 1
we get
L.H.S= ( 1 – sin^2 A cos^2 A )/( 2 + sin^2 A cos^2 A) = R.H.S 

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