Advertisements
Advertisements
Question
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Advertisements
Solution
Since, ABC is a right angled triangle, right angled at B.
So, A + C = 90°
`(secA.cosecC - tanA.cotC)/sinB`
= `(sec(90^circ - C).cosecC - tan(90^circ - C).cotC)/sin90^circ`
= `(cosecC.cosecC - cotC.cotC)/1`
= 1 ...[∵ cosec2θ – cot2θ = 1]
APPEARS IN
RELATED QUESTIONS
Write all the other trigonometric ratios of ∠A in terms of sec A.
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use tables to find sine of 47° 32'
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
Sin 2B = 2 sin B is true when B is equal to ______.
