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​cosec⁡2910∘+sec⁡4260∘+tan⁡2565∘+cot⁡1755∘=?\cosec2910^\circ + \sec 4260^\circ + \tan 2565^\circ + \cot 1755^\circ = ?cosec2910∘+sec4260∘+tan2565∘+cot
Question

cosec2910+sec4260+tan2565+cot1755=?\cosec2910^\circ + \sec 4260^\circ + \tan 2565^\circ + \cot 1755^\circ = ?​​

A.

4

B.

3

C.

0

D.

1

Correct option is A

Given:
cosec2910+sec4260+tan2565+cot1755\cosec 2910^\circ + \sec 4260^\circ + \tan 2565^\circ + \cot 1755^\circ​​
Concept Used:
Periodic Properties of Trigonometric Functions:
The period of cosecθ,secθ,tanθ,\cosec \theta, \sec \theta, \tan \theta,​ and cotθ\cot \theta​ is 360°.
Using the property f(θ+360)=f(θ)f(\theta + 360^\circ) = f(\theta)​​
for trigonometric functions, we will reduce the angles to their equivalent values within [0,360][0^\circ, 360^\circ]​​
Solution:
=cosec(2880+30)+sec(432060)+tan(2520+45)+cot(180045)=cosec30+sec60+tan45cot45=2+2+11=4= \cosec (2880 + 30)^\circ + \sec (4320 - 60)^\circ + \tan (2520 + 45)^\circ + \cot (1800 - 45)^\circ \\= \cosec 30^\circ + \sec 60^\circ + \tan 45^\circ - \cot 45^\circ \\= 2 + 2 + 1 - 1 \\= 4  
Thus, correct option is (a).

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