Advertisements
Advertisements
प्रश्न
Evaluate the following: `(sec32° cot26°)/(tan64° "cosec"58°)`
Advertisements
उत्तर
`(sec32° cot26°)/(tan64° "cosec"58°)`
= `(sec(90° - 58°) cot(90° - 64°))/(tan64° "cosec"58°)`
= `("cosec"58° cot64°)/(tan64° "cosec"58°)`
= 1.
APPEARS IN
संबंधित प्रश्न
If 4 cos2 x° - 1 = 0 and 0 ∠ x° ∠ 90°,
find:(i) x°
(ii) sin2 x° + cos2 x°
(iii) `(1)/(cos^2xx°) – (tan^2 xx°)`
Use the given figure to find:
(i) tan θ°
(ii) θ°
(iii) sin2θ° - cos2θ°
(iv) Use sin θ° to find the value of x.
Solve for x : cos2 30° + cos2 x = 1
If A = 30°, verify that cos2θ = `(1 - tan^2 θ)/(1 + tan^2 θ)` = cos4θ - sin4θ = 2cos2θ - 1 - 2sin2θ
In the given figure, if tan θ = `(5)/(13), tan α = (3)/(5)` and RS = 12m, find the value of 'h'.
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos84° + cosec69° - cot68°
Evaluate the following: cot20° cot40° cot45° cot50° cot70°
Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`
If sin(θ - 15°) = cos(θ - 25°), find the value of θ if (θ-15°) and (θ - 25°) are acute angles.
If A + B = 90°, prove that `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")` = tan2A
