The value of cos-sin 90- -sin - -sin 180- -tan 90- - is

The correct option is A 1 sin(90^∘ +θ)= cos θ sin(180^∘ +θ)= sin θ tan(90^∘ θ)= cot θ sin( θ)= sinθ Using these identities we get 1 ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

The value of ...

Question

The value of $\frac{cos Θ}{sin (90 + Θ)} + \frac{sin (- Θ)}{sin (180 + Θ)} + \frac{tan (90 + Θ)}{cot Θ}$ is

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Solution

The correct option is A 1
$s i n (90^{\circ} + θ) = c o s θ$
$s i n (180^{\circ} + θ) = - s i n θ$
$t a n (90^{\circ} - θ) = - c o t θ$
$s i n (- θ) = - s i n θ$
Using these identities we get 1

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Similar questions

Q. The value of :

\frac{c o s θ}{s i n (90 + θ)} + \frac{s i n (- θ)}{s i n (180 + θ)} + \frac{t a n (90 + θ)}{c o t θ}

Q. Evaluate

\frac{cos θ}{sin (90 + θ)} + \frac{sin (- θ)}{sin (180 + θ)} + \frac{tan (90 + θ)}{cot θ}

Q. Simplify

\frac{sec (90^{\circ} - θ)}{sin (90^{\circ} - θ)} \times \frac{cos θ}{tan (90^{\circ} - θ)} - sec θ

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. If

tan θ = \frac{5}{12}

and

θ

is not in the fourth quardrant then

\frac{tan (90^{\circ} + θ) - sin (180^{\circ} - θ)}{sin (270^{\circ} - θ) + c o s e c (360^{\circ} - θ)} =

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The value of cosΘsin(90+Θ)+sin(−Θ)sin(180+Θ)+tan(90+Θ)cotΘ is

The correct option is A 1sin(90∘+θ)=cosθsin(180∘+θ)=−sinθtan(90∘−θ)=−cotθsin(−θ)=−sinθUsing these identities we get 1

The value of $\frac{cos Θ}{sin (90 + Θ)} + \frac{sin (- Θ)}{sin (180 + Θ)} + \frac{tan (90 + Θ)}{cot Θ}$ is

The correct option is A 1
$s i n (90^{\circ} + θ) = c o s θ$
$s i n (180^{\circ} + θ) = - s i n θ$
$t a n (90^{\circ} - θ) = - c o t θ$
$s i n (- θ) = - s i n θ$
Using these identities we get 1