$\frac{cos (90 - A) sin (90 - A)}{tan (90 - A)}$ =

{sin}^{2} A

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{cos}^{2} A

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sin A

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1

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Solution

The correct option is A ${sin}^{2} A$
$= \frac{c o s (90 - A) s i n (90 - A)}{t a n (90 - A)}$
$= \frac{s i n A \times c o s A}{\frac{c o s A}{s i n A}} = \frac{s i n^{2} A \times c o s A}{c o s A}$
$= s i n^{2} A$

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Similar questions

Q. Prove the following :
(iv)

\frac{cos (90^{\circ} - A) sin (90^{\circ} - A)}{tan (90^{\circ} - A)} = {sin}^{2} A

Prove the following:

(i) $s i n θ s i n (90^{o} - θ) - c o s θ c o s (90^{o} - θ) = 0$
(ii) $\frac{c o s (90^{o} - θ) s e c (90^{o} - θ) t a n θ}{c o s e c (90^{o}) s i n (90^{o} - θ) c o t (90^{o} - θ)} + \frac{t a n (90^{o} - θ)}{c o t θ} = 2$
(iii) $\frac{t a n (90^{o} - A) c o t A}{c o s e c^{2} A} - c o s^{2} A = 0$
(iv) $\frac{c o s (90^{o} - A) s i n (90^{o} - A)}{t a n (90^{o} - A)} = s i n^{2} A$
(v) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$

Q. Prove that

\frac{cos (90 - A) \cdot sin (90 - A)}{tan (90 - A)} = {sin}^{2} A

\frac{sin 2 A}{sin A} - \frac{cos 2 A}{cos A} = sec A

Q. Show that

\frac{1 + sin 2 A}{cos 2 A} = \frac{cos A + sin A}{cos A - sin A} = cot (\frac{π}{4} - A)

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cos(90−A)sin(90−A)tan(90−A)=

The correct option is A sin2A=cos(90−A)sin(90−A)tan(90−A)=sinA×cosAcosAsinA=sin2A×cosAcosA=sin2A

$\frac{cos (90 - A) sin (90 - A)}{tan (90 - A)}$ =

The correct option is A ${sin}^{2} A$
$= \frac{c o s (90 - A) s i n (90 - A)}{t a n (90 - A)}$
$= \frac{s i n A \times c o s A}{\frac{c o s A}{s i n A}} = \frac{s i n^{2} A \times c o s A}{c o s A}$
$= s i n^{2} A$