Advertisements
Advertisements
प्रश्न
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
Advertisements
उत्तर
Since, A and B are complementary angles, A + B = 90°
cosec2 A + cosec2 B
= cosec2 A + cosec2 (90° – A)
= cosec2 A + sec2 A
= `1/sin^2A+1/cos^2A`
= `(cos^2A + sin^2A)/(sin^2Acos^2A)`
= `1/(sin^2Acos^2A)`
= cosec2 A sec2 A
= cosec2 A sec2 (90° – B)
= cosec2 A cosec2 B
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Prove that:
`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`
What is the maximum value of \[\frac{1}{\sec \theta}\]
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =
Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`
`tan 47^circ/cot 43^circ` = 1
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
