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
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
