Question #2fa3d

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Question #2fa3d

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Answer 1 · 2017-02-12T00:05:06

$\frac{- sec (x) {tan}^{2} (x) - (9 - sec (x)) {sec}^{2} (x)}{{tan}^{2} x}$ $(− sec ( x ) tan^2 ( x ) − ( 9 − sec ( x ) ) sec^2 ( x ))/tan^2 x$

$= \frac{- sec (x) ({sec}^{2} (x) - 1) - (9 - sec (x)) {sec}^{2} (x)}{{tan}^{2} x}$ $=(− sec ( x ) (sec^2 ( x ) -1)− ( 9 − sec ( x ) ) sec^2 ( x ))/tan^2 x$

$= \frac{- {sec}^{3} (x) + sec x - 9 {sec}^{2} x + {sec}^{3} (x)}{{tan}^{2} x}$ $=(− sec^3 ( x )+ secx− 9sec^2x + sec^3 ( x ))/tan^2 x$

$= \frac{sec x - 9 {sec}^{2} x}{{tan}^{2} x}$ $=(secx− 9sec^2x )/tan^2 x$

$= \frac{sec x}{{tan}^{2} x} - \frac{9 {sec}^{2} x}{{tan}^{2} x}$ $=secx/tan^2x− (9sec^2x)/tan^2 x$

$= cos x \times {csc}^{2} x - 9 {csc}^{2} x$ $=cosx xxcsc^2x− 9csc^2x$

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Question #2fa3d

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