Advertisements
Advertisements
प्रश्न
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
Advertisements
उत्तर
cos 45° = `1/sqrt(2)`
`(cos47^circ)/(sin43^circ) = (cos(90^circ - 43^circ))/(sin43^circ) = (sin43^circ)/(sin43^circ)` = 1 ...[cos (90 − θ) = sin θ]
`(sin72^circ)/(cos18^circ) = (cos(90^circ - 18^circ))/(cos18^circ) = (cos18^circ)/(cos18^circ)` = 1 ...[sin (90 − θ) = cos θ]
`((cos47^circ)/(sin43^circ))^2 + ((sin72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
= `1^2 + 1^2 - 2(1/sqrt(2))^2`
= `1 + 1 - 2(1/2)`
= 2 – 1
= 1
APPEARS IN
संबंधित प्रश्न
If tan A = cot B, prove that A + B = 90°.
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
If cot( 90 – A ) = 1, then ∠A = ?
The value of (tan1° tan2° tan3° ... tan89°) is ______.
