Advertisements
Advertisements
प्रश्न
If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`
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
Use the given figure to find:
(i) tan θ°
(ii) θ°
(iii) sin2θ° - cos2θ°
(iv) Use sin θ° to find the value of x.
Calculate the value of A, if (cosec 2A - 2) (cot 3A - 1) = 0
If sin 3A = 1 and 0 < A < 90°, find cos 2A
If 3 tan A - 5 cos B = `sqrt3` and B = 90°, find the value of A
Find the value 'x', if:
A ladder is placed against a vertical tower. If the ladder makes an angle of 30° with the ground and reaches upto a height of 18 m of the tower; find length of the ladder.
In the given figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically upwards and then 20 km at 60° to the vertical. PQ represents the first stage of the journey and QR the second. S is a point vertically below R on the horizontal level as P, find:
a. the height of the rocket when it is at point R.
b. the horizontal distance of point S from P.
Evaluate the following: `(sec34°)/("cosec"56°)`
If cos3θ = sin(θ - 34°), find the value of θ if 3θ is an acute angle.
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
