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

Evaluate:

sin 27° sin 63° – cos 63° cos 27°

Evaluate:

`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`

Evaluate:

3 cos 80° cosec 10° + 2 cos 59° cosec 31°

Evaluate:

`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`

Prove that:

tan (55° - A) - cot (35° + A)

Prove that:

sec (70° – θ) = cosec (20° + θ)

Prove that:

sin (28° + A) = cos (62° – A)

Prove that:

`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`

Prove that:

`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^2(90^@ - A)`

If A and B are complementary angles, prove that:

cot B + cos B = sec A cos B (1 + sin B)

If A and B are complementary angles, prove that:

cot A cot B – sin A cos B – cos A sin B = 0

If A and B are complementary angles, prove that:

cosec² A + cosec² B = cosec² A cosec² B

If A and B are complementary angles, prove that:

`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`

Find A, if 0° ≤ A ≤ 90° and 2 cos² A – 1 = 0

Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0

Find A, if 0° ≤ A ≤ 90° and 4 sin² A – 3 = 0

Find A, if 0° ≤ A ≤ 90° and cos² A – cos A = 0

Find A, if 0° ≤ A ≤ 90° and 2 cos² A + cos A – 1 = 0

If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`

Using remainder Theorem, factorise:

2x³ + 7x² − 8x – 28 Completely