The value of the expression tan180-cos180-sec-sec360-cot90-sin90- is

Given nbsp;tan(180° θ)cos(180°+θ)sec( θ)sec(360° θ)cot(90° θ)sin(90°+θ)Using nbsp;the nbsp;various nbsp;conditions nbsp;for nbsp;trigonometric nbs ...

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Standard XIII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

The value of ...

Question

$The value of the expression \frac{tan (180 ° - θ) cos (180 ° + θ) sec (- θ)}{sec (360 ° - θ) cot (90 ° - θ) sin (90 ° + θ)} is$

- 1

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\frac{1}{2}

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- 2

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Solution

The correct option is D $1$
$Given \frac{tan (180 ° - θ) cos (180 ° + θ) sec (- θ)}{sec (360 ° - θ) cot (90 ° - θ) sin (90 ° + θ)} Using the various conditions for trigonometric f unctions of allied angles \frac{tan (180 ° - θ) cos (180 ° + θ) sec (- θ)}{sec (360 ° - θ) cot (90 ° - θ) sin (90 ° + θ)} = \frac{(- tan θ) (- cos θ) (sec θ)}{sec θ (tan θ) (cos θ)} = 1$

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Similar questions

Q. The value of

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

is

(a) 1
(b) − 1
(c) 2
(d) −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

Prove that:

$(i) \frac{c o s (2 π + θ) c o s e c (2 π + θ) t a n (π / 2 + θ)}{s e c (π / 2 + θ) c o s θ c o t (π + θ)} = 1$

$(i i) \frac{c o s e c (90^{\circ} + θ) + c o t (450^{\circ} + θ)}{c o s e c (90^{\circ} - θ) + t a n (180^{\circ} - θ)} + \frac{t a n (180^{\circ} + θ) + s e c (180^{\circ} - θ)}{t a n (360^{\circ} + θ) - s e c (- θ)} = 2$

$(i i i) \frac{s i n (180^{\circ} + θ) c o s (90^{\circ} + θ) t a n (270^{\circ} - θ) c o t (360^{\circ} - θ)}{s i n (360^{\circ} - θ) c o s (360^{\circ} + θ) c o s e c (- θ) s i n (270^{\circ} + θ)} (i v) 1 + c o t θ - s e c (\frac{π}{2} + θ)} 1 + c o t θ + s e c (\frac{π}{2} + θ)} = 2 c o t θ$

$(v) \frac{t a n (90^{\circ} - θ) s e c (180^{\circ} - θ) s i n (- θ)}{s i n (180^{\circ} + θ) c o t (360^{\circ} - θ) c o s e c (90^{\circ} - θ)} = 1$

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

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

Q. Choose the correct answer from the alternatives given :
The value of

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

is equal to

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The value of the expression tan(180°−θ)cos(180°+θ)sec(−θ)sec(360°−θ)cot(90°−θ)sin(90°+θ) is

$The value of the expression \frac{tan (180 ° - θ) cos (180 ° + θ) sec (- θ)}{sec (360 ° - θ) cot (90 ° - θ) sin (90 ° + θ)} is$