Question

sinx=3/5,x lies in second quadrant,then find five trigonometry functions

Answer 1

sin x = 3/5

sin^2 x + cos^2 x = 1
(3/5)^2 + cos^2 x = 1
cos^2 x = 1 - 9/25
cos^2 x = 16/25
cos x = + or - 4/5
But x lies in the second quadrant, so
cos x = -4/5

cosec x = 1 / sin x
cosec x = 5/3

sec x = - 5/4

tan x = -3/4. and cot x =-4/3

Answer 2

Answer:

$cosx=-\frac{4}{5}, secx=-\frac{5}{4}, tanx=-\frac{3}{4}, cosecx=\frac{5}{3}, cotx=- \frac{4}{3}$

Step-by-step explanation:

Given: $sinx=\frac{3}{5}$ & $x$ lies in second quadrant.

To find all trigonometry functions.

We know that in second quadrant sine & cosine functions are positive rest are negative.

$sinx=\frac{P}{H}$

Using Pythagoras theorem: $H^{2}=P^{2}+B^{2}$.

Here, $P=3,H=5$ substituting we get, $B=4$.

So, all trigonometry functions are given as follows:

$cosx=\frac{B}{H} =-\frac{4}{5}$      

$tanx=\frac{P}{B}=-\frac{3}{4} \\cosecx=\frac{H}{P}=\frac{5}{3}\\ secx=\frac{H}{B}=-\frac{5}{4}\\ cotx=\frac{B}{P}=-\frac{4}{3}$

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