Advertisements
Advertisements
प्रश्न
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
Advertisements
उत्तर
`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
tan(49°) = cot(41°)
(because 49° + 41° = 90°)
`(cot^2 41°)/(tan^2 49°) = (cot^2 41°)/(cot41°)^2 = 1`
`(sin^2 75°)/(cos^2 15°)`
sin75° = cos15°
`(sin^2 75°)/(cos^2 15°) = (cos15°)^2/(cos15°)^2 = 1`
1 − 2(1) = 1 − 2
= −1
APPEARS IN
संबंधित प्रश्न
Show that cos 38° cos 52° − sin 38° sin 52° = 0
if `tan theta = 3/4`, find the value of `(1 - cos theta)/(1 +cos theta)`
Find the value of x, if cos (2x – 6) = cos2 30° – cos2 60°
Use tables to find sine of 21°
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
∠ACD is an exterior angle of Δ ABC. If ∠B = 40o, ∠A = 70o find ∠ACD.
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If A and B are complementary angles, then
Solve: 2cos2θ + sin θ - 2 = 0.
If sec A + tan A = x, then sec A = ______.
