Advertisements
Advertisements
प्रश्न
In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.
Advertisements
उत्तर
sin θ = `("AB")/("AC")` and tan θ = `("AB")/("BC")`
APPEARS IN
संबंधित प्रश्न
In Given Figure, find tan P – cot R.
If cot θ =` 7/8` evaluate `((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`
State whether the following are true or false. Justify your answer.
cot A is the product of cot and A.
Prove that `(sin "A" - 2sin^3 "A")/(2cos^3 "A" - cos "A") = tan "A"`
If `cos θ = 12/13`, show that `sin θ (1 - tan θ) = 35/156`.
If `cot theta = 1/sqrt3` show that `(1 - cos^2 theta)/(2 - sin^2 theta) = 3/5`
If `tan θ = 20/21` show that `(1 - sin theta + cos theta)/(1 + sin theta + cos theta) = 3/7`
Evaluate the Following
4(sin4 60° + cos4 30°) − 3(tan2 60° − tan2 45°) + 5 cos2 45°
Evaluate the Following
`4/(cot^2 30^@) + 1/(sin^2 60^@) - cos^2 45^@`
Evaluate the Following:
`(tan^2 60^@ + 4 cos^2 45^@ + 3 sec^2 30^@ + 5 cos^2 90)/(cosec 30^@ + sec 60^@ - cot^2 30^@)`
Find the value of x in the following :
`sqrt3 sin x = cos x`
If `sqrt2 sin (60° – α) = 1` then α is ______.
The value of sin² 30° – cos² 30° is ______.
The value of the expression `[(sin^2 22^circ + sin^2 68^circ)/(cos^2 22^circ + cos^2 68^circ) + sin^2 63^circ + cos 63^circ sin 27^circ]` is ______.
What will be the value of sin 45° + `1/sqrt(2)`?
`sqrt(3)` cos2A + `sqrt(3)` sin2A is equal to ______.
Prove that `tan θ/(1 - cot θ) + cot θ/(1 - tanθ)` = 1 + sec θ cosec θ
Find an acute angle θ when `(cos θ - sin θ)/(cos θ + sin θ) = (1 - sqrt(3))/(1 + sqrt(3))`
In ΔBC, right angled at C, if tan A = `8/7`, then the value of cot B is ______.
