90- sin270- cosec180-cos - sin90- tan360- is equal to

The correct option is B θ cosec θ nbsp;(90° θ)sin(270°+θ)cosec(180°+θ)cos( θ)sin(90°+θ)tan(360° θ) =tanθ( cosθ)( cosec θ)(cosθ)(cosθ)( tanθ) = θ cosec θ n ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

90∘-θ sin270∘...

Question

$\frac{cot (90 ° - θ) sin (270 ° + θ) cosec (180 ° + θ)}{cos (- θ) sin (90 ° + θ) tan (360 ° - θ)}$ is equal to

sec θ cosec θ

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- sec θ cosec θ

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- cosec θ

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sec θ

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Solution

The correct option is B $- sec θ cosec θ$
$\frac{cot (90 ° - θ) sin (270 ° + θ) cosec (180 ° + θ)}{cos (- θ) sin (90 ° + θ) tan (360 ° - θ)}$
$= \frac{tan θ (- cos θ) (- cosec θ)}{(cos θ) (cos θ) (- tan θ)}$
$= - sec θ cosec θ$

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Similar questions

\frac{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)} =

Q. Solve

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

Q. Prove that
(i)

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

(ii)

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

(iii)

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

(iv)

\{1 + cotθ - \sec (\frac{π}{2} + θ)\} \{1 + cotθ + \sec (\frac{π}{2} + θ)\} = 2 cotθ

(v)

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

$\frac{c o t (90^{o} - θ) s i n (90^{o} - θ)}{s i n θ} + \frac{c o t 40^{o}}{t a n 50^{o}} - (c o s^{2} 20^{o} + c o s^{2} 70^{o})$ = ?

(a) 0 (b) 1 (c) -1 (d) none of these

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

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Sign of Trigonometric Ratios in Different Quadrants

Standard XII Mathematics

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cot(90°−θ)sin(270°+θ)cosec(180°+θ)cos(−θ)sin(90°+θ)tan(360°−θ) is equal to

The correct option is B −secθ cosec θ cot(90°−θ)sin(270°+θ)cosec(180°+θ)cos(−θ)sin(90°+θ)tan(360°−θ) =tanθ(−cosθ)(−cosec θ)(cosθ)(cosθ)(−tanθ) =−secθ cosec θ

$\frac{cot (90 ° - θ) sin (270 ° + θ) cosec (180 ° + θ)}{cos (- θ) sin (90 ° + θ) tan (360 ° - θ)}$ is equal to

The correct option is B $- sec θ cosec θ$
$\frac{cot (90 ° - θ) sin (270 ° + θ) cosec (180 ° + θ)}{cos (- θ) sin (90 ° + θ) tan (360 ° - θ)}$
$= \frac{tan θ (- cos θ) (- cosec θ)}{(cos θ) (cos θ) (- tan θ)}$
$= - sec θ cosec θ$