1-sin90- - -cos290- - -cos90- -1-sin90-

1+sin(90^∘ θ) cos^2(90^∘ θ)/cos(90^∘ θ)[1+sin(90^∘ θ)] =1+cos θ sin^2 θ/sin θ(1+cos θ) =(1 sin^2 θ)+cos θ/sin θ (1+cos θ) =cos^2 θ + cos θ/sin θ(1+cos θ ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

1+sin90∘ - θ-...

Question

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

$cot θ$

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$tan θ$

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Solution

The correct option is A
$cot θ$

$\frac{1 + sin (90^{\circ} - θ) - {cos}^{2} (90^{\circ} - θ)}{cos (90^{\circ} - θ) [1 + sin (90^{\circ} - θ)]}$

$= \frac{1 + cos θ - {sin}^{2} θ}{sin θ (1 + cos θ)}$

$= \frac{(1 - {sin}^{2} θ) + cos θ}{sin θ (1 + cos θ)}$

$= \frac{{cos}^{2} θ + cos θ}{sin θ (1 + cos θ)}$

$= \frac{cos θ (cos θ + 1)}{sin θ (1 + cos θ)} = cot θ$

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Similar questions

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

Q. Prove that

\frac{sin (90^{\circ} - θ)}{1 + sin θ} + \frac{cos θ}{1 - cos (90^{\circ} - θ)} = 2 sec θ

Q. Evaluate:

\frac{cos θ}{sin (90^{\circ} - θ)} + \frac{sin θ}{cos (90^{\circ} - θ)}

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

Find the value of
$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]

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1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ) [1+sin(90∘−θ)]=

The correct option is A cot θ 1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ)[1+sin(90∘−θ)] =1+cos θ−sin2 θsin θ(1+cos θ) =(1−sin2 θ)+cos θsin θ(1+cos θ) =cos2 θ+cos θsin θ(1+cos θ) =cos θ(cos θ+1)sin θ(1+cos θ)=cot θ

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