Solving LHS, tan^2ϕ sin^2ϕ = gt; #160; sin^2ϕcos^2ϕ sin^2ϕ = gt; #160; sin^2ϕ(1cos^2ϕ 1) = gt; #160; sin^2ϕ(sec^2ϕ 1) = gt; # ...

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Complementary Trigonometric Ratios

tan2ϕ-sin2ϕ=t...

Question

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

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Solution

Solving LHS,

$t a n^{2} ϕ - s i n^{2} ϕ$

=> $\frac{s i n^{2} ϕ}{c o s^{2} ϕ} - s i n^{2} ϕ$

=> $s i n^{2} ϕ (\frac{1}{c o s^{2} ϕ} - 1)$

=> $s i n^{2} ϕ (s e c^{2} ϕ - 1)$

=> $s i n^{2} ϕ . t a n^{2} ϕ$ = RHS

Hence proved,

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