Advertisements
Advertisements
Question
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
Advertisements
Solution
`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
RELATED QUESTIONS
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Solve.
`cos22/sin68`
Solve.
sin42° sin48° - cos42° cos48°
Evaluate.
`(cos^2 32^@+cos^2 58^@)/(sin^2 59^@+sin^2 31^@)`
Evaluate.
`(sin77^@/cos13^@)^2+(cos77^@/sin13^@)-2cos^2 45^@`
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
The value of \[\frac{\tan 55°}{\cot 35°}\] + cot 1° cot 2° cot 3° .... cot 90°, is
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
