Advertisements
Advertisements
प्रश्न
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Advertisements
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.
संबंधित प्रश्न
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
sec4 A − sec2 A is equal to
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
If 1 – cos2θ = `1/4`, then θ = ?
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]`
