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प्रश्न
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
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उत्तर
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A [[ cos A" sin A " ],[-sin A" cos A"]]`
= ` [[sin^2A " - sin A cos A"],[sinA .cos A - sin^2 A ]]+ [[cos^2 A " cos A . sin A"],[ -sinA cos A cos^2 A]]`
` =[[sin^2 A + cos^2 A " - sin A. cos A + cos A . sin A "],[sin A . cos A - sin A . cos A " sin^2 A + cos^2 A]] = [[ 1 0 ] , [ 0 1]]`
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संबंधित प्रश्न
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
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`
