Advertisements
Advertisements
प्रश्न
Evaluate 2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 45°.
उत्तर
2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ
= `2(sqrt(2))^2 + 3(sqrt(2))^2 - 2. 1/sqrt(2). 1/sqrt(2)` ...[∵ θ = 45°]
= `2 xx 2 + 3 xx 2 - 2 xx 1/2`
= 4 + 6 – 1
= 9
APPEARS IN
संबंधित प्रश्न
If cot θ =` 7/8` evaluate `((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
tan θ = 11
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
`tan alpha = 5/12`
If sin θ = `12/13`, Find `(sin^2 θ - cos^2 θ)/(2sin θ cos θ) × 1/(tan^2 θ)`.
if `cot theta = 3/4` prove that `sqrt((sec theta - cosec theta)/(sec theta +cosec theta)) = 1/sqrt7`
Evaluate the following
tan2 30° + tan2 60° + tan2 45°
Evaluate the Following
cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
= `square/square + square/square` ......`[∵ cot θ = square/square, tan θ = square/square]`
= `(square + square)/(square xx square)` .....`[∵ square + square = 1]`
= `1/(square xx square)`
= `1/square xx 1/square`
= cosec θ·sec θ ......`[∵ "cosec" θ = 1/square, sec θ = 1/square]`
= R.H.S.
∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec·sec θ
Find an acute angle θ when `(cos θ - sin θ)/(cos θ + sin θ) = (1 - sqrt(3))/(1 + sqrt(3))`
If sinθ = `1/sqrt(2)` and `π/2 < θ < π`. Then the value of `(sinθ + cosθ)/tanθ` is ______.
