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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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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Solution

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

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Similar questions

Q. 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

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

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

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

is equal to

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

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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

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