Advertisements
Advertisements
प्रश्न
Evaluate:
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
Advertisements
उत्तर
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
`2 tan(90^circ - 33^circ)/(cot33^circ) - cot(90^circ - 20^circ)/(tan20^circ) - sqrt(2)(1/sqrt2)`
`2 cot33^circ/(cot33^circ) - tan20^circ/(tan20^circ) - 1`
= 2 – 1 – 1
= 0
संबंधित प्रश्न
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate cosec 31° − sec 59°
Solve.
sin42° sin48° - cos42° cos48°
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Evaluate:
14 sin 30° + 6 cos 60° – 5 tan 45°
Prove that:
`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`
Without using trigonometric tables, prove that:
sec70° sin20° + cos20° cosec70° = 2
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
The value of 3 sin 70° sec 20° + 2 sin 49° sec 51° is
