Question

sin x=3/5,x lies in second quadrant,find value of cos x and tan x​

Answer 1

Answer:-

• Sinx = 3/5
• x lies in second quadrant

Since x lies in 2nd quadrant the value of x is positive.

Firstly we calculate the value of cos x .

As we know relationship between sin & cos

• sin²x + cos²x = 1

substitute the value we get

$:\implies$ (3/5)² + cos²x = 1

$:\implies$ 9/25 + cos²x = 1

$:\implies$ cos²x = 1-9/25

$:\implies$ cos²x = 25-9/25

$:\implies$ cos²x = 16/25

$:\implies$ cosx = ± √16/25

$:\implies$ cosx = ±4/5

Since the lie in second quadrant then the value of x is in negative

• Therefore , cosx = -4/5

Now, calculating the value of tanx

As we know that ,

$:\implies$ tanx = sinx/cosx

Substitute the value we get

$:\implies$ tanx = 3/5 ÷ -4/5

$:\implies$ tanx = 3/5 × -5/4

$:\implies$ tanx = -3/4

Since x lie in 2nd quadrant so, the value of x is negative.

• Therefore , tanx = -(-3/4) = 3/4

• Hence, the value of cosx & tanx are -4/5 & 3/4 respectively.

Answer 2

Step-by-step explanation:

Given:-

• sin x=3/5
• x lies in second Quadrant .

To find:-

Cos x and tan x

Solution:-

As x lies in second Quadrant cos x and tan x will be negative.

we know that

$\boxed{\sf cos\: x=\sqrt{1-sin^2x}}$

$\\ \tt{:}\Rrightarrow \sqrt{1-(3/5)^2}$

$\\ \tt{:}\Rrightarrow \sqrt{1-\dfrac{9}{25}}$

$\\ \tt{:}\Rrightarrow \sqrt{\dfrac{25-9}{25}}$

$\\ \tt{:}\Rrightarrow \sqrt{\dfrac{16}{25}}$

$\\ \tt{:}\Rrightarrow \underline{+}\dfrac{4}{5}$

• Cos x will be negative so cos x=-4/5

Now

$\boxed{\sf tan\:x=\dfrac{sin\:x}{cos\:x}}$

$\\ \tt{:}\Rrightarrow \dfrac{\dfrac{3}{5}}{\dfrac{-4}{5}}$

$\\ \tt{:}\Rrightarrow \dfrac{-3}{4}$

• tan x will be negative

$\\ \bf{:}\Rrightarrow tanx=-\dfrac{3}{4}$