Advertisements
Advertisements
प्रश्न
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Advertisements
उत्तर
LHS = (sin2θ)2 + (cos2 θ)2 + 2 sin2θ cos2θ - 2 sin2θ cos2θ
= ( sin2θ + cos2θ )2 - 2 sin2θ cos2θ
= 1 - 2 sin2θ cos2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove that:
`cosA/(1 + sinA) = secA - tanA`
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
(sec θ + tan θ) . (sec θ – tan θ) = ?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
