Advertisements
Advertisements
प्रश्न
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
पर्याय
0
1
`1/2`
`sqrt(3)/2`
उत्तर
If ∆ABC is right angled at C, then the value of cos (A + B) is 0.
Explanation:
We know that,
In ∆ABC,
Sum of three angles = 180°
i.e., ∠A + ∠B + ∠C = 180°
But right angled at C
i.e., ∠C = 90° ...[Given]
∠A + ∠B + 90° = 180°
⇒ A + B = 90° ...[∵∠A = A]
∴ cos (A + B) = cos 90° = 0
संबंधित प्रश्न
If the angle θ = -60° , find the value of sinθ .
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
Evaluate.
sin(90° - A) cosA + cos(90° - A) sinA
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
Prove that:
sec (70° – θ) = cosec (20° + θ)
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
A, B and C are interior angles of a triangle ABC. Show that
If ∠A = 90°, then find the value of tan`(("B+C")/2)`
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
The value of tan 72° tan 18° is
Choose the correct alternative:
If ∠A = 30°, then tan 2A = ?
`tan 47^circ/cot 43^circ` = 1
