Question
Sun July 22, 2012 By: Chethan U

# If SinA=2pq/p[square}+q[square] Pt SecA-Tan A=p-q/p+q

Sun July 22, 2012
Answer : Given :SinA=2pq / (p2+q2)
To Prove : SecA - Tan A= (p - q ) / (p+q)

Now ,
=> SinA = 2pq / (p2+q2)
Using componendo and dividendo,we get
=> ( sinA +1 ) / (sinA -1) = ( 2pq + p2 + q2 ) / (p2 + q2 - 2pq )
Inverting the values we get ,
=> ( sinA - 1 ) / (sinA + 1) =  (p2 + q2 - 2pq ) / ( 2pq + p2 + q2)
=> ( sinA - 1 ) / (sinA + 1) = (p - q)2 / ( p + q)2
rationalising LHS
=> (1- sinA)2 / ( 1- sin2A ) = (p - q)2 / ( p + q)2
=> (1- sinA)/ cos2A = (p - q)2 / ( p + q)2     { using sin2A + cos2A =1 ,
=> 1- sin2A = cos2A }
taking the squareroot both the sides , we get
=> (1- sinA) / cosA = (p - q) / ( p + q)
=>(1/ cosA) - (sinA / cosA) = (p - q) / ( p + q )
=> secA - tanA = (p - q) / ( p + q)        { using secA = 1/cosA and                                                                           tanA = sinA/cosA }
Hence Proved

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Thu December 14, 2017

# I am sending the question and solution of the same sent by toppers today. It appears from the solution part the formula used is totally wrong. Because in a triangle how sum of the squares of any two sides is less than the square of the third side ? It is wrong assumption and therefore the whole solution is based on wrong assumption. I therefore request Toppers to scrutinize the solutions before forwarding the same to the users  and please do not put the users in confusion.

Thu December 14, 2017