Advertisements
Advertisements
प्रश्न
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Advertisements
उत्तर
`cos38^circ sec(90^circ - 2A) = 1`
⇒ `cos38^circcosec2A = 1`
⇒ `cos38^circ (1/(sin2A)) = 1`
⇒ `sin2A = cos(90 - 52^circ)`
⇒ `sin2A = sin52^circ`
⇒ `2A = 52^circ`
⇒ `A = 26^circ`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Prove that sec2θ – cos2θ = tan2θ + sin2θ
