(English Medium) ICSE Class 10 - CISCE Question Bank Solutions for Mathematics

Prove that sin⁴θ - cos⁴θ = sin²θ - cos²θ
= 2sin²θ - 1
= 1 - 2 cos²θ

Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.

`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A

Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.

Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`

Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`

Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`

Prove that ( 1 + tan A)² + (1 - tan A)² = 2 sec²A

If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x² - y² = a² - b².

Prove that: 2(sin⁶θ + cos⁶θ) - 3 ( sin⁴θ + cos⁴θ) + 1 = 0.

Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .

Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.

If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.

Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`

Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.

Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.

Prove that sec² (90° - θ) + tan² (90° - θ) = 1 + 2 cot² θ.

Prove that cosec² (90° - θ) + cot² (90° - θ) = 1 + 2 tan² θ.

Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.

Prove that sin( 90° - θ ) sin θ cot θ = cos²θ.