Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 25 - Complementary Angles [Latest edition]

Evaluate : `(cos22°)/(sin 68°)`

Evaluate: `(tan 47°)/(cot 43°)`

Evaluate: `(sec 75°)/("cosec" 15°)`

Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`

Evaluate: sin² 40° - cos² 50°

Evaluate: sec² 18° - cosec² 72°

Evaluate: sin 15° cos 15° - cos 75° sin 75°

Evaluate: sin 42° sin 48° - cos 42° cos 48°

Evaluate: sin (90° - A) sin A - cos (90° - A) cos A

Evaluate: sin² 35° - cos² 55°

Evaluate: `(cot54°)/(tan36°) + (tan20°)/(cot70°)-2`

Evaluate: `(2tan53°)/(cot 37°)-(cot 80°)/(tan 10°)`

Evaluate: cos² 25° - sin² 65° - tan² 45°

Evaluate: `((sin 77°)/(cos 13°))^2 + ((cos 77°)/(sin 13°))^2 - 2 cos^2 45°`

Show that: tan 10° tan 15° tan 75° tan 80° = 1.

Show that: sin 42° sec 48° + cos 42° cosec 48° = 2.

Express the following in term of angles between 0° and 45° :

sin 59° + tan 63°

Express the following in term of angles between 0° and 45° :

cosec 68° + cot 72°

Express the following in term of angles between 0° and 45° :

cos 74° + sec 67°

For triangle ABC, show that:

`sin (("A + B")/(2)) = cos  "C"/(2)`

For triangle ABC, show that:

`tan(("B"+"C")/2) = cot("A"/2)`

Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.

Evaluate:

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

Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°

Evaluate:

tan (55° – A) – cot (35° + A)

Evaluate: cosec (65° + A) - sec (25° - A)

Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`

Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`

Evaluate: `(cos70°)/(sin20°) + (cos59°)/(sin 31°) - 8 sin^2 30°`

Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.

A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.

In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.

In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.

cos(90° - A) · sec 77° = 1