Advertisements
Advertisements
प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Advertisements
उत्तर
In ∆ABC, we have
A + B + C = 180º
⇒ B + C = 180º – A
`\Rightarrow \frac{B+C}{2}=\text{ }90^\text{o}-\frac{A}{2}`
Taking tan on both sides, we get
`\Rightarrow \tan ( \frac{B+C}{2})=\tan( 90^\text{o}-\frac{A}{2})`
`\Rightarrow \tan ( \frac{B+C}{2} )=\cot \frac{A}{2}`
संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Solve.
`cos55/sin35+cot35/tan55`
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Find the value of x, if cos (2x – 6) = cos2 30° – cos2 60°
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
The value of tan 10° tan 15° tan 75° tan 80° is
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
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)`
If sin 3A = cos 6A, then ∠A = ?
