Question

p.t. costheta / 1- sintheta
= 1+sintheta / costheta​

Answer 1

$\mathcal\red{Question:-}$

p.t. cos∅ / 1- sin∅

= 1+sin∅ / cos∅

$\mathcal\red{Solution:-}$

We have ,

LHS

= cos∅ / 1- sin∅

Multiplying numerator and denominator by ( 1+ sin∅)

= cos∅(1+sin∅) / (1-sin∅)(1+sin∅)

= cos∅ (1+sin∅) / 1-sin^2∅

{ °.° (a-b) (a-b) = a^2 - b^2 }

= cos∅ ( 1+sin∅) / cos^2∅

{ °.° 1- sin^2∅ = cos^2∅ }

= 1+sin∅ / cos∅

= RHS

Hence proved

$\bold\red{Thank\:You}$❤

Answer 2

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