Advertisements
Advertisements
प्रश्न
Find A, if 0° ≤ A ≤ 90° and cos2 A – cos A = 0
Advertisements
उत्तर
cos2 A – cos A = 0
`=>` cos A (cos A – 1) = 0
`=>` cos A = 0 or cos A = 1
We know cos 90° = 0 and cos 0° = 1
Hence, A = 90° or 0°
APPEARS IN
संबंधित प्रश्न
Write all the other trigonometric ratios of ∠A in terms of sec A.
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
The value of tan 1° tan 2° tan 3° ...... tan 89° is
The value of
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
The value of \[\frac{\tan 55°}{\cot 35°}\] + cot 1° cot 2° cot 3° .... cot 90°, is
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
If tan θ = 1, then sin θ . cos θ = ?
