Advertisements
Advertisements
प्रश्न
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Advertisements
उत्तर
Consider the table.
| θ | 0° | 30° | 45° | 60° | 90° |
| `sin θ` | `0` | `1/2` | `1/sqrt2` | `sqrt3/2` | `1` |
| `cosθ` | `1` | `sqrt3/2` | `1/sqrt2` | `1/2` | `0` |
Here,
`sin 90°+cos 90°=1+0=1` Which is not greater than 1 Therefore, the given statement is false,
APPEARS IN
संबंधित प्रश्न
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Find the value of ( sin2 33° + sin2 57°).
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
If tan θ × A = sin θ, then A = ?
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
