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
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Use tables to find cosine of 2° 4’
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Use tables to find the acute angle θ, if the value of cos θ is 0.9574
Prove that:
tan (55° - A) - cot (35° + A)
Find A, if 0° ≤ A ≤ 90° and cos2 A – cos A = 0
Sin 2A = 2 sin A is true when A =
Find the sine ratio of θ in standard position whose terminal arm passes through (4,3)
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
