Advertisements
Advertisements
प्रश्न
The value of cosec(70° + θ) – sec(20° − θ) + tan(65° + θ) – cot(25° − θ) is
विकल्प
0
1
2
3
Advertisements
उत्तर
0
Explanation;
Hint:
cosec(70° + θ) – sec(20° – θ) + tan(65° + θ) – cot(25° – θ)
= sec[90° – (70° + θ)] – sec(20° – θ) + tan(65° + θ) – tan[90° – (25° – θ)]
= sec(20° – θ) – sec(20° – θ) + tan(65° + θ) – tan(65° + θ)
= 0
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
If A and B are complementary angles, prove that:
cot A cot B – sin A cos B – cos A sin B = 0
∠ACD is an exterior angle of Δ ABC. If ∠B = 40o, ∠A = 70o find ∠ACD.
If \[\tan A = \frac{5}{12}\] \[\tan A = \frac{5}{12}\] find the value of (sin A + cos A) sec A.
The value of \[\frac{\cos^3 20°- \cos^3 70°}{\sin^3 70° - \sin^3 20°}\]
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
