Advertisements
Advertisements
प्रश्न
Prove that:
sec (70° – θ) = cosec (20° + θ)
बेरीज
Advertisements
उत्तर
sec (70° – θ) = sec [90° – (20° + θ)] = cosec (20° + θ)
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Trigonometrical Identities - Exercise 21 (E) [पृष्ठ ३३२]
APPEARS IN
संबंधित प्रश्न
Solve.
sin15° cos75° + cos15° sin75°
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Evaluate:
`cos70^circ/(sin20^circ) + cos59^circ/(sin31^circ) - 8 sin^2 30^circ`
Use tables to find sine of 62° 57'
Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
If sin 3A = cos 6A, then ∠A = ?
Prove that `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`.
