The value of - cos -sin90-sin-sin180-tan90-cot -

The correct option is B 1 cos(90 θ)=sinθ sin(90 θ)=cosθ tan(90 θ)=θ (90 θ)=tanθ (90 θ)=θ cosec(90 θ)=θ cosθsin(90+θ)+sin( θ)s ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

The value of ...

Question

The value of :
$\frac{c o s θ}{s i n (90 + θ)} + \frac{s i n (- θ)}{s i n (180 + θ)} + \frac{t a n (90 + θ)}{c o t θ}$

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Solution

The correct option is B 1
$cos (90 - θ) = sin θ$
$sin (90 - θ) = cos θ$
$tan (90 - θ) = cot θ$
$cot (90 - θ) = tan θ$
$sec (90 - θ) = csc θ$
$c o s e c (90 - θ) = sec θ$

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

$= \frac{c o s θ}{c o s (θ)} + \frac{- s i n (θ)}{- s i n (θ)} - \frac{c o t (θ)}{c o t θ}$

$= 1 + 1 - 1$

$= 1$

Hence answer is $B$

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Similar questions

Q. The value of

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

The correct option is B 1cos(90−θ)=sinθsin(90−θ)=cosθtan(90−θ)=cotθcot(90−θ)=tanθsec(90−θ)=cscθcosec(90−θ)=secθcosθsin(90+θ)+sin(−θ)sin(180+θ)+tan(90+θ)cotθ=cosθcos(θ)+−sin(θ)−sin(θ)−cot(θ)cotθ=1+1−1=1Hence answer is B

The value of :
$\frac{c o s θ}{s i n (90 + θ)} + \frac{s i n (- θ)}{s i n (180 + θ)} + \frac{t a n (90 + θ)}{c o t θ}$