Question

Sin^2 63° + sin^2 27° + cos^2 17° + cos^2 73°

Answer 1

Answer:

= sin²63° + sin²27° + cos²17°+ cos²73°

= sin²90 + cos²90

= 1²+0²

= 1

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Answer 2

${\huge{\boxed{\sf{\green{❥✰Question✰}}}}}$

$:-\sf \dfrac{ {sin}^{2} 63 + {sin}^{2}27 }{cos {}^{2} 17 + {cos}^{2} 73} cos 2 17+cos 2 73sin 2 63+sin 2 27$

${\huge{\boxed{\sf{\red{❥✰Answer✰}}}}}$

Given :-

$\begin{gathered}\longmapsto \sf\bold{\pink{\dfrac{{sin}^{2}63 + {sin}^{2}27}{{cos}^{2}17 + {cos}^{2}73}}}\\\end{gathered} ⟼ cos 2 17+cos 2 73sin 2 63+sin 2 27$

Solution :-

$\leadsto \sf \dfrac{{sin}^{2}63 + {sin}^{2}27}{{cos}^{2}17 + {cos}^{2}73}⇝ cos 2 17+cos 2 73sin 2 63+sin 2 27$

$\implies \sf \dfrac{{sin}^{2}63 + {sin}^{2}(90 - 63)}{{cos}^{2}17 + {cos}^{2}(90 - 17)}⟹ cos 2 17+cos 2 (90−17)sin 2 63+sin 2 (90−63)$

As we know that,

$\mapsto \sf sin(90 - \theta) =\: cos\theta↦sin(90−θ)=cosθ$

$\mapsto \sf cos(90 - \theta) =\: sin\theta↦cos(90−θ)=sinθ$

$\implies \sf \dfrac{{sin}^{2}63 + {cos}^{2}63}{{cos}^{2}17 + {sin}^{2}17}⟹ cos 2 17+sin 2 17sin 2 63+cos 2 63$

Again, as we know that,

$\mapsto \sf {sin}^{2}A + {cos}^{2}A =\: 1↦sin 2 A+cos 2 A=1$

$\implies \sf \dfrac{\cancel{1}}{\cancel{1}}⟹ 1​ 1$

$\implies \sf\boxed{\bold{\red{1}}}⟹ 1$

$\sf\boxed{\bold{\green{\therefore\: The\: answer\: is\: 1.}}}$

$\rule{150}{2}$

∴The answer is 1.

Some Important Formula :

$mapsto \sf\bold{\purple{cos(90 - \theta) =\: sin\theta}}↦cos(90−θ)=sinθ\mapsto \sf\bold{\purple{ sec(90 - \theta) =\: cosec\theta}}↦sec(90−θ)=cosecθ\mapsto \sf\bold{\purple{tan(90 - \theta) =\: cot\theta}}↦tan(90−θ)=cotθ\mapsto \sf\bold{\purple{cosec(90 - \theta) =\: sec\theta}}↦cosec(90−θ)=secθ\mapsto \sf\bold{\purple{sin(90 - \theta) =\: cos\theta}}↦sin(90−θ)=cosθ\mapsto \sf\bold{\purple{cot(90 - \theta) =\: tan\theta}}↦cot(90−θ)=tanθ$

Thanks!