Advertisements
Advertisements
Question
If cos θ = `24/25`, then sin θ = ?
Solution
cos θ = `24/25` ......[Given]
We know that,
sin2θ + cos2θ = 1
∴ `sin^2theta + (24/25)^2` = 1
∴ `sin^2theta + 576/625` = 1
∴ sin2θ = `1 - 576/625`
∴ sin2θ = `(625 - 576)/625`
∴ sin2θ = `49/625`
∴ sin θ = `7/25` ......[Taking square root of both sides]
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
` tan^2 theta - 1/( cos^2 theta )=-1`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
If tanθ `= 3/4` then find the value of secθ.
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
