If tanA+sinA=p and tanA-sinA=q. Then proove that #p^2-q^2=4sqrt (pq)# ?

3 Answers
Swarna Islam · Sunayan S
Dec 28, 2017

It is proved below:

Explanation:

Here,
#tanA+sinA=p#

#tanA-sinA=q#

#L.H.S=p^2-q^2#

#=(tanA+sinA)^2-(tanA-sinA)^2#

#=(tanA+cancelsinA+tanA-cancelsinA)(canceltanA+sinA-canceltanA+sinA)#

#=(2tanA)cdot(2sinA)#

#=4tanAsinA#

#=4sqrt(tan^2Asin^2A)#

#=4sqrt((sec^2A-1)sinA)##color(blue)([[As,sec^2theta-tan^2theta=1]])#

#=4sqrt(sec^2A.sin^2A-sin^2A)#

#=4sqrt(sin^2A/(cos^2A)-sin^2A)#

#=4sqrt(tan^2A-sin^2A)#

#=4sqrt((tanA+sinA)(tanA-sinA))#

#=4sqrt(pq)##color(green)([As.tanA+sinA=pandtanA-sinA=q])#

#=R.H.S#

Ratnaker Mehta
Dec 30, 2017

Kindly refer to the Explanation.

Explanation:

#tanA+sinA=p......(1), and, tanA-sinA=q......(2)#.

#:. (1)+(2) rArr 2tanA=p+q, or, tanA=(p+q)/2#.

#"Similarly, "sinA=(p-q)/2#.

#:. cotA=2/(p+q), and cscA=2/(p-q)#.

But, #csc^2A-cot^2A=1#,

#rArr 4/(p-q)^2-4/(p+q)^2=1, or, #.

#[4{(p+q)^2-(p-q)^2}]/{(p+q)^2(p-q)^2}=1#.

# rArr (4*4pq)/(p^2-q^2)^2=1, i.e., #

# (p^2-q^2)^2=16pq#,

# rArr p^2-q^2=4sqrt(pq),# as desired!

Enjoy Maths.!

P dilip_k
Dec 30, 2017

#RHS=4sqrt(pq)#

#=4sqrt((tanA+sinA)(tanA-sinA))#

#=4sqrt(tan^2A-sin^2A)#

#=4sqrt(sin^2A/cos^2A-sin^2A)#

#=4sqrt(sin^2Axxsec^2A-sin^2A)#

#=4sqrt(sin^2A(sec^2A-1)#

#=4sqrt(sin^2Atan^2A)#

#=4tanAsinA#

#=(tanA+sinA)^2-(tanA-sinA)^2#

#=p^2-q^2=RHS#

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