Advertisements
Advertisements
प्रश्न
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Advertisements
उत्तर
LHS = `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ`
= `(sin^4θ + cos^4θ)/(sin^2θ.cos^2θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2(sin^2 θ. cos^2 θ))/(sin^2 θ.cos^2 θ)`
= `((1)^2 - 2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - (2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - 2`
= RHS
संबंधित प्रश्न
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Prove that `sec"A"/(tan "A" + cot "A")` = sin A
Prove that sec2θ – cos2θ = tan2θ + sin2θ
If cos A + cos2A = 1, then sin2A + sin4 A = ?
