Advertisements
Advertisements
प्रश्न
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Advertisements
उत्तर
LHS = `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2`
⇒ `(1 + sin^2 θ + cos^2 θ + 2(sin θ - cos θ - sin θ. cos θ))/(1 + sin^2 θ + cos^2 θ + 2(sin θ + cos θ + sin θ. cos θ)`
= `(1 + 1 + 2 (sin θ - cos θ - sin θ. cos θ))/( 1 + 1 + 2((sin θ + cos θ + sin θ. cos θ)`
= `(2 (1 + sin θ - cos θ - sin θ. cos θ))/(2( 1 + (sin θ + cos θ + sin θ. cos θ))`
= `( 1 + sin θ - cos θ( 1 + sin θ))/(1 + sin θ + cos θ( 1 + sin θ))`
= `((1 + sin θ)(1 - cos θ))/((1 + sin θ)( 1 + cos θ))`
= `(1 - cos θ)/( 1 + cos θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
If sec θ = `25/7`, then find the value of tan θ.
tan θ cosec2 θ – tan θ is equal to
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
