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प्रश्न
If `cos theta [(cos theta, sin theta),(-sin theta, cos theta)] + sin theta[(x, -cos theta),(cos theta, x)]` = I2, find x.
उत्तर
`[(cos theta, sin theta),(-sin theta, cos theta)] xx [(cos theta, - x),(x, cos theta)]` = I2
`[(cos^2theta + xsintheta, -xsintheta + sinthetacostheta),(-sinthetacostheta + xcostheta, xsintheta + cos^2theta)] = [(1, 0),(0, 1)]`
`[(cos^2theta + xsintheta, 0),(0, xsintheta + cos^2theta)] = [(1, 0),(0, 1)]`
cos2θ + x sinθ = 1
x sinθ = 1 – cos2θ
x sinθ = sin2θ
x = `(sin^2theta)/(sin theta)`
x = sinθ
– x cosθ + sinθ cosθ = 0
sinθ cosθ = x cosθ
⇒ sinθ = x ...(÷ cosθ)
∴ x = sinθ
The value of x = sinθ
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