Advertisements
Advertisements
प्रश्न
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
उत्तर
LHS = `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2`
= `((sin 20°. sin 70°)/(cos 20°))^2 + ((cos 20°. cos 20°)/(sin 20°))^2`
= `[(sin 20°.sin (90° - 20°))/(cos 20°)]^2 + [(cos 20°. cos(90° - 20°))/(sin 20°)]^2`
= `[ (sin 20°.cos 20°)/(cos 20°)]^2 + [(cos 20°. sin 20°)/(sin 20°)]^2`
= sin2 20° + cos2 20°
= 1
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`