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Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Concept: undefined >> undefined
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
Concept: undefined >> undefined
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Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
Concept: undefined >> undefined
If x = `θ/360` × 2πr then what is x in the formula?
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In the given figure, a rectangle ABCD is inscribed inside a semi-circle of radius 10 cm. Using the dimensions given in the figure, determine the area of the shaded region.
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The perimeter of an arc of radius 4.2 cm is 12.8 cm. Determine the angle subtended by the arc at the centre of circle.
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In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.
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In the adjoining figure, a tangent is drawn to a circle of radius 4 cm and centre C, at the point S. Find the length of the tangent ST, if CT = 10 cm.
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If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
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In an equilateral triangle PQR, prove that PS2 = 3(QS)2.
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Concept: undefined >> undefined
Eliminate θ if x = r cosθ and y = r sinθ.
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`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
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Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
Concept: undefined >> undefined
In the following figure, m(arc PMQ) = 130o, find ∠PQS.
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In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle if m(arc PCR) = 26°, m(arc QDS) = 48°, then find:
(i) m∠PQR
(ii) m∠SPQ
(iii) m∠RAQ
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Write the equation of the line passing through A(–3, 4) and B(4, 5) in the form of ax + by + c = 0
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Show that points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are vertices of a rhombus ABCD.
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If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
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If \[\tan \theta = \frac{3}{4}\], find the values of secθ and cosθ
Concept: undefined >> undefined
