Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Advertisements
उत्तर
We have to prove `tan theta - cot theta = (2 sin^2 theta - `1)/(sin theta cos theta)`
We know that. `sin^2 theta + cos^2 theta - 1`
So,
`tan theta - cot theta = sin theta/cos theta - cos theta/sin theta`
`= (sin^2 theta - cos^2 theta)/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (2 sin^2 theta - 1)/(sin theta cos theta)`
APPEARS IN
संबंधित प्रश्न
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
From the figure find the value of sinθ.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A . "cosec" A + 1`.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
