In right angled ABC- - B - 90- and - A - - then cos 90 - - -

The correct option is A sinθ In the figure, cos (90 θ) = yr and sinθ = yr So, cos (90 θ) = sinθ So, sinθ is correct. ...

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Standard XII

Mathematics

Property 1

In right angl...

Question

In right angled $△ A B C, ∠ B = 90^{\circ}$ and $∠ A = θ$ , then $cos (90 - θ) =$

sin θ

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tan θ

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cot θ

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None of above

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Solution

The correct option is A $sin θ$
In the figure,
$cos (90 - θ) = \frac{y}{r}$ and $sin θ = \frac{y}{r}$

So, $cos (90 - θ) = sin θ$

So, $sin θ$ is correct.

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Similar questions

Q. In a triangle ABC right angled at B,

∠ A C B = θ .

If t a n θ = \frac{5}{12}

then find the value of \frac{(1 + c o s θ) \times (1 - s i n θ)}{(1 + s i n θ) \times (1 - c o s θ)} .

Q. Prove:

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

Q. Evaluate:

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

Q. If

tan θ = \frac{a}{b}

, then

\frac{sin θ}{{cos}^{8} θ} + \frac{cos θ}{{sin}^{8} θ} =

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In right angled △ABC,∠B=90∘ and ∠A=θ, then cos(90−θ)=

The correct option is A sinθIn the figure,cos(90−θ)=yr and sinθ=yrSo, cos(90−θ)=sinθSo, sinθ is correct.

In right angled $△ A B C, ∠ B = 90^{\circ}$ and $∠ A = θ$ , then $cos (90 - θ) =$

The correct option is A $sin θ$
In the figure,
$cos (90 - θ) = \frac{y}{r}$ and $sin θ = \frac{y}{r}$

So, $cos (90 - θ) = sin θ$

So, $sin θ$ is correct.