Advertisements
Advertisements
प्रश्न
If θ is an acute angle and sin θ = cos θ, find the value of tan2 θ + cot2 θ – 2.
उत्तर
Given, sin θ = cos θ
`sinθ/cosθ` = 1
tan θ = 45°
`\implies` θ = 45°
So tan2 θ + cot2 θ – 2 = tan2 45° + cot2 45° – 2
= 1 + 1 – 2
= 0
∴ tan2 θ + cot2 θ – 2 = 0
APPEARS IN
संबंधित प्रश्न
If sin A = `3/4`, calculate cos A and tan A.
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`cosec theta = sqrt10`
if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`
Evaluate the following
cos2 30° + cos2 45° + cos2 60° + cos2 90°
Evaluate the Following
4(sin4 60° + cos4 30°) − 3(tan2 60° − tan2 45°) + 5 cos2 45°
Evaluate the Following
4(sin4 30° + cos2 60°) − 3(cos2 45° − sin2 90°) − sin2 60°
Find the value of x in the following :
`sqrt3 tan 2x = cos 60^@ + sin45^@ cos 45^@`
In ΔABC is a right triangle such that ∠C = 90° ∠A = 45°, BC = 7 units find ∠B, AB and AC
`(1 + tan^2 "A")/(1 + cot^2 "A")` is equal to ______.
If cosec θ = `("p" + "q")/("p" - "q")` (p ≠ q ≠ 0), then `|cot(π/4 + θ/2)|` is equal to ______.