Cos90- cosec270- cos180- - sin270- sin360-

The correct option is A 1cos(90°+θ)cosec(270°+θ)cos(180°+θ)( θ)sin(270°+θ)sin(360° θ) = sinθ( θ)( cosθ)θ( cosθ)( sinθ)= 1 ...

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Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

cos90∘+θ cose...

Question

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

- 1

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1

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{cot}^{2} θ

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tan θ

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Solution

The correct option is A $- 1$
$\frac{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)}$

$= \frac{- sin θ (- sec θ) (- cos θ)}{sec θ (- cos θ) (- sin θ)} = - 1$

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Similar questions

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

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:

\frac{cos (90 - θ)}{sin θ} + \frac{sin θ}{cos (90 - θ)} = 2

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

is equal to

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

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Trigonometric Ratios of Allied Angles

Standard XII Mathematics

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cos(90°+θ)cosec(270°+θ)cos(180°+θ)sec(−θ)sin(270°+θ)sin(360°−θ)=

The correct option is A −1cos(90°+θ)cosec(270°+θ)cos(180°+θ)sec(−θ)sin(270°+θ)sin(360°−θ) =−sinθ(−secθ)(−cosθ)secθ(−cosθ)(−sinθ)=−1

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

The correct option is A $- 1$
$\frac{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)}$

$= \frac{- sin θ (- sec θ) (- cos θ)}{sec θ (- cos θ) (- sin θ)} = - 1$