$tan^{2} ϕ - sin^{2} ϕ = tan^{2} ϕ . sin^{2} ϕ$

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tan^{2} ϕ - sin^{2} ϕ = tan^{2} ϕ . sin^{2} ϕ

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Text solutionVerified

Solving LHS,

t a n^{2} ϕ - s i n^{2} ϕ

\frac{s i n ^{2} ϕ}{co s ^{2} ϕ} - s i n^{2} ϕ

s i n^{2} ϕ (\frac{1}{co s ^{2} ϕ} - 1)

s i n^{2} ϕ (se c^{2} ϕ - 1)

s i n^{2} ϕ . t a n^{2} ϕ

= RHS

Hence proved,

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