Prove that - 1 - sin 2 A - 1 - sin 2 A - tan - 4 - A

√(1+sin 2 A1 sin 2 A)=√(sin^2 A+cos^2 A+2sin Acos Asin^2 A+cos^2 A 2sin Acos A)=√((sin A+cos A)^2(cos A sin A)^2)=sin A+cos Acos A sin A=1+tan A1 tan A=tan ...

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Byju's Answer

Standard XII

Mathematics

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

Prove that √...

Question

Prove that $\sqrt{\frac{1 + sin 2 A}{1 - sin 2 A}} = tan (\frac{π}{4} + A)$

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Solution

$\sqrt{\frac{1 + sin 2 A}{1 - sin 2 A}} = \sqrt{\frac{{sin}^{2} A + {cos}^{2} A + 2 sin A cos A}{{sin}^{2} A + {cos}^{2} A - 2 sin A cos A}} = \sqrt{\frac{(sin A + cos A)^{2}}{(cos A - sin A)^{2}}} = \frac{sin A + cos A}{cos A - sin A} = \frac{1 + tan A}{1 - tan A} = tan (\frac{π}{4} + A)$

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Prove that√1+sin2A1−sin2A=tan(π4+A)

√1+sin2A1−sin2A=√sin2A+cos2A+2sinAcosAsin2A+cos2A−2sinAcosA=√(sinA+cosA)2(cosA−sinA)2=sinA+cosAcosA−sinA=1+tanA1−tanA=tan(π4+A)

Prove that $\sqrt{\frac{1 + sin 2 A}{1 - sin 2 A}} = tan (\frac{π}{4} + A)$

$\sqrt{\frac{1 + sin 2 A}{1 - sin 2 A}} = \sqrt{\frac{{sin}^{2} A + {cos}^{2} A + 2 sin A cos A}{{sin}^{2} A + {cos}^{2} A - 2 sin A cos A}} = \sqrt{\frac{(sin A + cos A)^{2}}{(cos A - sin A)^{2}}} = \frac{sin A + cos A}{cos A - sin A} = \frac{1 + tan A}{1 - tan A} = tan (\frac{π}{4} + A)$