sin(90 -theta) . sin theta / tan theta - 1 = -sin ^2 theta
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Solution -
Given: secθ + tanθ = p
We know that, tan^2θ + 1 = sec^2θ
Therefore, sec^2θ - tan^2θ = 1
(secθ + tanθ) (secθ - tanθ) = 1
p (secθ - tanθ) = 1
secθ - tanθ = 1/p
Hence proved.
Given: secθ + tanθ = p
We know that, tan^2θ + 1 = sec^2θ
Therefore, sec^2θ - tan^2θ = 1
(secθ + tanθ) (secθ - tanθ) = 1
p (secθ - tanθ) = 1
secθ - tanθ = 1/p
Hence proved.
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