Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Advertisements
उत्तर १
LHS = (tan θ + sec θ)2 + (tan θ - sec θ)2
`"LHS" = tan^2 θ + sec^2 θ + cancel(2 tan θ. sec θ) + tan^2 θ + sec^2 θ - cancel(2 tan θ. sec θ) ...{(a^2 + b^2 = a^2 + 2ab + b^2),(a^2 - b^2 = a^2 - 2ab + b^2):}`
LHS = `2 tan^2θ + 2 sec^2θ`
LHS = `2[tan^2θ + sec^2θ]`
LHS = `2[sin^2 θ/cos^2 θ + 1/cos^2 θ]`
LHS = `2((sin^2 θ + 1)/cos^2 θ)`
LHS = `2((1 + sin^2 θ)/(1 - sin^2θ)) ...{(sin^2θ + cos^2θ = 1),(∴ cos^2θ = 1 - sin^2θ):}`
RHS = `2((1 + sin^2 θ)/(1 - sin^2θ))`
LHS = RHS
उत्तर २
LHS = `[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 ...{(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)}`
LHS = `2[tan^2θ + 1/cos^2θ]`
LHS = `2[sin^2 θ/cos^2 θ + 1/cos^2 θ]`
LHS = `2((sin^2 θ + 1)/cos^2 θ)`
LHS = `2((1 + sin^2 θ)/(1 - sin^2θ)) ...{(sin^2θ + cos^2θ = 1),(∴ cos^2θ = 1 - sin^2θ):}`
RHS = `2((1 + sin^2 θ)/(1 - sin^2θ))`
LHS = RHS
APPEARS IN
संबंधित प्रश्न
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
(secA + tanA) (1 − sinA) = ______.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
If tan θ = `13/12`, then cot θ = ?
If 1 – cos2θ = `1/4`, then θ = ?
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
