Advertisements
Advertisements
प्रश्न
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
Advertisements
उत्तर
Since, A and B are complementary angles, A + B = 90°
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA)`
= `(sinA + sinB)/(sinA - sinB) + (cos(90^@ - A) - cos(90^@ - B))/(cos(90^@ - A) + cos(90^@ - B))`
= `(sinA + sinB)/(sinA - sinB) + (sinA - sinB)/(sinA + sinB)`
= `((sinA + sinB)^2 + (sinA - sinB)^2)/((sinA - sinB)(sinA + sinB)`
= `(sin^2A + sin^2B + 2sinAsinB + sin^2A + sin^2B - 2sinA)/(sin^2A - sin^2B`
= `2(sin^2A + sin^2B)/(sin^2A - sin^2B)`
= `2(sin^2A + sin^2(90^@ - A))/(sin^2A - sin^2(90^@ - A))`
= `2(sin^2A + cos^2B)/(sin^2A - cos^2B)`
= `2/(sin^2A - (1 - sin^2A))`
= `2/(2sin^2A - 1)`
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Evaluate.
cos225° + cos265° - tan245°
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Find the sine ratio of θ in standard position whose terminal arm passes through (3, 4)
What is the maximum value of \[\frac{1}{\sec \theta}\]
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
The value of tan 1° tan 2° tan 3° ...... tan 89° is
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
In ∆ABC, `sqrt(2)`AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ?, ∠B = ?, ∠C = ?
