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Question
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
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Solution
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^@73^@)`
`(= [sin(90^@ - 27^@)]^2+sin^2 27^@)/([cos(90^@ - 73^@)]^2 + cos^2 73^@)`
`= ([cos27^@]^2 + sin^2 27^@)/([sin 73^@]^2 + cos^2 73^@)`
`= (cos^2 27^@ + sin^2 27^@)/(sin^2 73^@+ cos^2 73^@)`
= 1/1 (As sin2A + cos2A = 1)
= 1
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cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
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Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
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.
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Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
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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
