Advertisements
Advertisements
प्रश्न
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
Advertisements
उत्तर
`sinA/(secA - 1) + sinA/(secA + 1) = 2`
`=> (sinAsecA + sinA + secAsinA - sinA)/((secA - 1)(secA + 1)) = 2`
`=> (2sinAsecA)/(sec^2A - 1) = 2`
`=> (sinAsecA)/tan^2A = 1`
`=> cosA/sinA = 1`
`=>` cot A = 1
We know cot 45° = 1
Hence, A = 45°
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Write all the other trigonometric ratios of ∠A in terms of sec A.
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
Evaluate.
sin235° + sin255°
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
Find the value of the following:
sin 21° 21′
