Advertisements
Advertisements
प्रश्न
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Advertisements
उत्तर
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
`sin(50^circ + θ) = cos[90^circ - (50^circ + θ)] = cos(40^circ - θ)`
`sin(50^circ + θ) - cos(40^circ - θ)`
= `cos(40^circ - θ) - cos(40^circ - θ)`
= 0
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
If `tan θ = 13/12`, then cot θ = ?
