Question

1) If ∠P + ∠Q = 90°, let's us show that, $\sqrt{\dfrac{sinP}{cosQ} - sinP cosQ = cosP}$
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Answer 1

Answer:

SOLUTION :-

$\leadsto$ L.H.S :- $\sqrt{\dfrac{sinP}{cosQ} - sinP cosQ}$

=> $\sqrt{\dfrac{sinP}{cos(90° - P)}sinP . cos(90° - p)}$

[ $\angle$P + $\angle$Q = 90° => $\angle$Q = 90° - $\angle$P ]

=> $\sqrt{\dfrac{sinP}{sinP}sinP . sinP}$

=> $\sqrt{1 - {sin}^{2} P}$

=> $\sqrt{{cos}^{2}P}$

=> cosP

L.H.S = R.H.S

$\mapsto$ $\boxed{\bold{\large{PROVED}}}$

Answer 2

$\color{red}\huge{\underline{\underline{GIVEN\dag}}}$

• ∠P + ∠Q = 90°

$\color{red}\huge{\underline{\underline{SOLUTION\dag}}}$

$\sqrt{\dfrac{sinP}{cosQ} - sinP cosQ }= cosP$

But,We know that;

$∠Q =90°-∠P$

$\sqrt{\dfrac{sinP}{cos(90 - P)} - sinP cos(90 - P)}$

We know that;

$\color{blue}{\boxed{\cos(90-θ)=\sinθ}}$

$= \sqrt{\dfrac{sinP}{ sinP} - sinP \times sin P }$

We also know that;

$\color{blue}{\boxed{1-\sin^{2}θ=\cos^{2}θ}}$

$= \sqrt{1 - \sin ^{2} ( P) }$

$= \sqrt{ \cos {}^{2} (P) }$

$\color{orange} = \cos( P)$

Therefore,

L.H.S.=R.H.S.

HOPE THAT HELPS!!