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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 that:
sec2θ + cosec2θ = sec2θ x cosec2θ
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
