Question

sin thetha + sec thetha whole square+ cos thetha+ cosec thetha whole square= 1 + sec thetha + cosec thetha whole square​

Answer 1

solution of the above problem

Answer 2

$\huge\red{\underline{\mathfrak{Solution}}}$

Taking LHS

→ ( sinθ + secθ)² +( cosθ + cosecθ)²

$= {(sinθ + \frac{1}{cosθ} )}^{2} + {(cosθ + \frac{1}{sinθ}) }^{2}$

$= ( { \frac{sinθ.cosθ + 1}{cosθ} })^{2} + {( \frac{sinθ.cosθ + 1}{sinθ} })^{2}$

$= ({sinθ.cosθ + 1 )}^{2} ( \frac{1}{ {cos}^{2} θ} + \frac{1}{ {sin}^{2}θ } )$

$= ( {sin}^{2}θ {cos}^{2} θ + 2sinθ.cosθ + 1)( \frac{ {sin}^{2} θ + {cos}^{2} θ}{{sin}^{2} θ . {cos}^{2} } )$

$= ( \frac{ {sin}^{2} θ . {cos}^{2} θ}{{sin}^{2} θ . {cos}^{2} } ) + \frac{2 \sinθ \cosθ }{ {sin}^{2} θ. { \ \cos }^{2} θ} + \frac{1}{ {sin}^{2}θ. { \cos}^{2}θ }$

$= 1 + 2secθ.cosθ + \: {sec}^{2}θ. {cosec}^{2}θ$

$= ({1 + secθ.cosecθ)}^{2}$

$= RHS$