Advertisements
Advertisements
प्रश्न
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove the following:
`(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Advertisements
उत्तर
LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`
= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`
= `(sec θ + 1)/(sec θ - 1)`
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
If cos \[9\theta\] = sin \[\theta\] and \[9\theta\] < 900 , then the value of tan \[6 \theta\] is
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
