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Question

If x is an acute angle and tan x=17, then the value of cosec2 x-sec2 xcosec2 x+sec2 x is
(a) 3/4
(b) 1/2
(c) 2
(d) 5/4

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Solution

(a) 3/4

We have:tan x = 17 tan2 x =17Now, dividing the numerator and the denominator of cosec2 x-sec2 xcosec2 x+sec2 x by cosec2 x:
1-tan2x1+tan2x= 1-171+17= 68 = 34

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