Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
उत्तर
LHS = `(cosA + sinA)^2 + (cosA - sinA)^2`
= `cos^2A + sin^2A + 2cosA.sinA + cos^2A + sin^2A - 2cosA.sinA`
= `2(cos^2A + sin^2A) = 2` = RHS
APPEARS IN
संबंधित प्रश्न
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
What is the value of (1 + cot2 θ) sin2 θ?
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ
If 2sin2θ – cos2θ = 2, then find the value of θ.
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
