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प्रश्न
` tan^2 theta - 1/( cos^2 theta )=-1`
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उत्तर
LHS= `tan^2 theta - 1/(cos^2 theta)`
=` (sin^2 theta )/( cos^2 theta) - 1/(cos^2 theta)`
=`(sin ^2 theta-1)/(cos^2 theta)`
=` (-cos^2 theta )/(cos^2 theta)`
= -1
= RHS
APPEARS IN
संबंधित प्रश्न
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Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
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`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
