Advertisements
Advertisements
प्रश्न
Prove the following: sin58° sec32° + cos58° cosec32° = 2
Advertisements
उत्तर
L.H.S.
= sin58° sec32° + cos58° cosec32°
= `sin(90° - 32°) xx (1)/(cos32°) + cos(90° - 32°) xx (1)/(sin32°)`
= `cos32° xx (1)/(cos32°) + sin32° xx (1)/(sin32°)`
= 1 + 1
= 2
= R.H.S.
APPEARS IN
संबंधित प्रश्न
State for any acute angle θ whether tan θ increases or decreases as θ decreases.
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Solve for x : cos `(x)/(3) –1` = 0
Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`
Find the length of EC.
If tan x° = `(5)/(12) . tan y° = (3)/(4)` and AB = 48m; find the length CD.
Evaluate the following: `(tan42°)/(cot48°) + (cos33°)/(sin57°)`
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: tan77° - cot63° + sin57°
If secθ= cosec30° and θ is an acute angle, find the value of 4 sin2θ - 2 cos2θ.
Prove the following: sin230° + cos230° = `(1)/(2)sec60°`
