Prove that- cos Asin90-A-sin Acos90-A-2

⇒ cosAsin(90 #176; A ) + #160;sinAcos(90 #176; A ) = 2 Taking L.H.S.= ⇒ cosAsin(90 #176; A ) + sinAcos(90 #176; A ) ⇒ ...

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Standard IX

Mathematics

Complementary Ratios

Prove that: ...

Question

Prove that:
$\frac{cos A}{sin (90^{\circ} - A)} + \frac{sin A}{cos (90^{\circ} - A)} = 2$

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Solution

$\Rightarrow$ $\frac{cos A}{sin (90 ° - A)} + \frac{sin A}{cos (90 ° - A)} = 2$

Taking L.H.S.=
$\Rightarrow$ $\frac{cos A}{sin (90 ° - A)} + \frac{sin A}{cos (90 ° - A)}$

$\Rightarrow$ $\frac{cos A}{cos A} + \frac{sin A}{sin A} [∵ cos (90 ° - θ) = sin θ and sin (90 ° - θ) = cos θ]$

$\Rightarrow$ $1 + 1$

$\Rightarrow$ $2$

$= R . H . S .$
Hence proved.

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Similar questions

$\frac{sin A}{sin (90 - A)} + \frac{cos A}{cos (90 - A)} = s e c A c o s e c A$

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\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{2 {sin}^{2} A - 1}

Prove that cosA sin $(90^{\circ} - A)$ = 1 - $s i n^{2} A$

Prove that:

(i) $s i n θ c o s (90^{\circ} - θ) + s i n (90^{\circ} - θ) c o s θ = 1$

(ii) $\frac{s i n θ}{c o s (90^{\circ} - θ)} + \frac{c o s θ}{s i n (90^{\circ} - θ)} = 2$

(iii) $\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1$

(iv) $\frac{c o s (90^{\circ} - θ) s e c (90^{\circ} - θ) t a n θ}{c o s e c (90^{\circ} - θ) s i n (90^{\circ} - θ) c o t (90^{\circ} - θ)} + \frac{t a n (90^{\circ} - θ)}{c o t θ} = 2$

(v) $\frac{c o s (90^{\circ} - θ)}{1 + s i n (90^{\circ} - θ)} + \frac{1 + s i n (90^{\circ} - θ)}{c o s (90^{\circ} - θ)} = 2 c o s e c θ$

(vi) $\frac{s e c (90^{\circ} - θ) c o s e c θ - t a n (90^{\circ} - θ) c o t θ + c o s^{2} 25^{\circ} + c o s^{2} 65^{\circ}}{3 t a n 27^{\circ} t a n 63^{\circ}} = \frac{2}{3}$

(vii) $c o t θ t a n (90^{\circ} - θ) - s e c (90^{\circ} - θ) c o s e c θ + \sqrt{3} t a n 12^{\circ} t a n 60^{\circ} t a n 78^{\circ} = 2$

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Prove that:cos Asin(90∘−A)+sin Acos(90∘−A)=2

⇒cosAsin(90°−A)+sinAcos(90°−A)=2Taking L.H.S.=⇒cosAsin(90°−A)+sinAcos(90°−A)⇒cosAcosA+sinAsinA[∵cos(90°−θ)=sinθ and sin(90°−θ)=cosθ]⇒1+1⇒2=R.H.S.Hence proved.

Prove that:
$\frac{cos A}{sin (90^{\circ} - A)} + \frac{sin A}{cos (90^{\circ} - A)} = 2$