Question

simplify cosec ( 2 sec^-1 3x + pi/2 ) in terms of x

Answer 1

help ?

Answer 2

sorry, the jack has very little clue
i can give u an answer, but not how i arrived at it?

Answer 3

okay , at least you're helping

Answer 4

\[ \large \csc ( 2 sec^{-1} (3x) + \frac {\pi}{2}) \]
\[ \large =\sec ( 2 sec^{-1} (3x)) \]
\[ \huge = \frac {1}{ \frac {2}{9 x^2}-1}\]
\[ \large = \frac {-9 x^2}{(9 x^2-2)}\]
you'll have to ask @ganeshie8 or @hartnn to help out with the "why" if they're free ???

Answer 5

@hartnn can you explain please ?

Answer 6

by the way thanks jack

Answer 7

\[\csc\left(2\sec^{-1}(3x)+\frac\pi2\right)\]\[=\sec\left[\frac\pi2-\left(2\sec^{-1}(3x)+\frac\pi2\right)\right]\]\[=\sec\left(-2\sec^{-1}(3x)\right)\]\[=-\sec\left(2\sec^{-1}(3x)\right)\]\[=-\sec\left(2\sec^{-1}(3x)\right)\]

Answer 8

can you explain it to me ?

Answer 9

\[=-\frac1{\cos\left(2\sec^{-1}(3x)\right)}\]\[=-\frac1{\cos^2\left(\sec^{-1}(3x)\right)-\sin^2\left(\sec^{-1}(3x)\right)}\]
Let \(\theta\) be \(\sec^{-1}(3x)\), \(\sec\theta=3x\), \(\cos\theta=\dfrac1{3x}\).
\[=-\frac1{\cos^2\theta-\sin^2\theta}\]\[=-\frac1{2\cos^2\theta-1}\]\[=-\frac1{\dfrac2{9x^2}-1}\]\[=-\frac1{\dfrac{2-9x^2}{9x^2}}\]\[=\frac{9x^2}{9x^2-2}\]

Answer 10

Because \(\sin\theta=\cos\left(\dfrac\pi2-\theta\right)\)

Answer 11

sorry gtg cya

Answer 12

thanks , that's helping a lot !

Answer 13

that was masterfully done dude! @kc_kennylau
props ;D

Answer 14

wow thanks for all the medals :D