Advertisements
Advertisements
प्रश्न
Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0
उत्तर
sin 3A – 1 = 0
`=>` sin 3A = 1
We know sin 90° = 1
∴ 3A = 90°
Hence, A = 30°
APPEARS IN
संबंधित प्रश्न
Evaluate cosec 31° − sec 59°
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
Prove that:
tan (55° - A) - cot (35° + A)
What is the maximum value of \[\frac{1}{\sec \theta}\]
If \[\tan A = \frac{5}{12}\] \[\tan A = \frac{5}{12}\] find the value of (sin A + cos A) sec A.
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
