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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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