Advertisements
Advertisements
प्रश्न
If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`
Advertisements
उत्तर
Since P, Q and R are interior angles of ΔPQR,
P + Q + R = 180°
⇒ Q + R = 180° - P
Now,
L.H.S. = `cot (("Q" + "R")/2)`
= `cot ((180° - "P")/2)`
= `cot(90° - "P"/2)`
= `tan "P"/(2)`
= R.H.S.
APPEARS IN
संबंधित प्रश्न
If 4 cos2 x° - 1 = 0 and 0 ∠ x° ∠ 90°,
find:(i) x°
(ii) sin2 x° + cos2 x°
(iii) `(1)/(cos^2xx°) – (tan^2 xx°)`
Solve for x : cos (2x - 30°) = 0
If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ
If `sqrt(3)`sec 2θ = 2 and θ< 90°, find the value of θ
Find the value 'x', if:
In the given figure; ∠B = 90°, ∠ADB = 30°, ∠ACB = 45° and AB = 24 m. Find the length of CD.
Evaluate the following: `(sec34°)/("cosec"56°)`
Evaluate the following: sin35° sin45° sec55° sec45°
Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`
Prove the following: sin58° sec32° + cos58° cosec32° = 2
