Advertisements
Advertisements
प्रश्न
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
Advertisements
उत्तर
Since Δ ABC is a right angled triangle, right angled at B,
A + C = 90°
∴ `(sec "A".sin "C" - tan "A". tan "C")/sin "B"`
= `(sec "A"(90° - "C")sin "C" - tan(90° - "C")tan "C")/(sin 90°)`
= `("cosec" "C" sin "C" - cot "C" tan "C")/(1)`
= `(1)/sin "C" xx sin "C" - (1)/tan "C" xx tan "C"`
= 1 - 1
= 0
APPEARS IN
संबंधित प्रश्न
Evaluate cosec 31° − sec 59°
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
Solve.
`cos22/sin68`
Evaluate:
`cos70^circ/(sin20^circ) + cos59^circ/(sin31^circ) - 8 sin^2 30^circ`
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
Write the maximum and minimum values of sin θ.
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
If 8 tan x = 15, then sin x − cos x is equal to
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
Evaluate: cos2 25° - sin2 65° - tan2 45°
