Prove the following -iv cos90- - A sin 90- - Atan90- - A - sin2 A

L.H.S =cos(90^∘ A) sin (90^∘ A)tan(90^∘ A) We have , #160; #160; tanA=1A =cos(90^∘ A) sin (90^∘ A) (90^∘ A) We ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Prove the fol...

Question

Prove the following :
(iv) $\frac{cos (90^{\circ} - A) sin (90^{\circ} - A)}{tan (90^{\circ} - A)} = {sin}^{2} A$

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Solution

L.H.S $= \frac{cos (90^{\circ} - A) sin (90^{\circ} - A)}{tan (90^{\circ} - A)}$

We have , $tan A = \frac{1}{cot A}$

$= cos (90^{\circ} - A) sin (90^{\circ} - A) cot (90^{\circ} - A)$

We know that,
$cos (90 - θ) = sin θ$
$sin (90 - θ) = cos θ$
$cot (90 - θ) = tan θ$

$∴ cos (90^{\circ} - A) sin (90^{\circ} - A) cot (90^{\circ} - A)$

$= sin (A) cos (A) t a n (A)$

$= \frac{sin (A) cos (A) s i n (A)}{cos (A)}$ --------------( $tan A = \frac{sin (A)}{cos A}$ )

$= sin (A) sin (A)$

$= {sin}^{2} (A)$ =R..H.S

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

Hence proved

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

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

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\frac{\cos (90 ° - θ) \sec (90 ° - θ) \tan θ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{\cot θ} = 2

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\frac{\tan (90 ° - A) \cot A}{{cosec}^{2} A} - \cos^{2} A = 0

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

(v) sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

Q. Prove the following :

\frac{cos (90^{\circ} - θ) sec (90^{\circ} - θ) tan θ}{cosec (90^{\circ} - θ) sin (90^{\circ} - θ) cot (90^{\circ} - θ)} + \frac{tan (90^{\circ} - θ)}{cot θ} = 2

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 the following :(iv) cos(90∘−A)sin(90∘−A)tan(90∘−A)=sin2A

Prove the following :
(iv) $\frac{cos (90^{\circ} - A) sin (90^{\circ} - A)}{tan (90^{\circ} - A)} = {sin}^{2} A$