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Question
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
Options
1
0
`1/2`
`sqrt2`
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Solution
sin θ × cosec θ = 1
Explanation:
sin θ × cosec θ
= `sintheta xx 1/sinθ ... [cosec theta = 1/(sintheta)]`
= 1
RELATED QUESTIONS
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
Prove that sec2θ − cos2θ = tan2θ + sin2θ
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
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`
