This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.
Solution:
Taking LHS;
= $\dfrac{1+sin2A -cos2A}{1+sin2A+cos2A}$
[ cos2A = 1- 2sin²A ]
= $\dfrac{1+sin2A -(1-2sin²A)}{1+sin2A+1-2sin²A}$
= $\dfrac{1+sin2A -1+2sin²A}{2+sin2A-2sin²A}$
[ 1 = sin²A +cos²A ]
= $\dfrac{sin2A +2sin²A}{2(1)+sin2A-2sin²A}$
= $\dfrac{sin2A +2sin²A}{2(sin²A +cos²A)+sin2A-2sin²A}$
= $\dfrac{sin2A +2sin²A}{2sin²A +2cos²A+sin2A-2sin²A}$
= $\dfrac{sin2A +2sin²A}{sin2A+2cos²A}$
[ sin2A = 2sinAcosA ]
= $\dfrac{2sinAcosA +2sin²A}{2sinAcosA+2cos²A}$
= $\dfrac{2sinA(cosA +sinA)}{2cosA(sinA+cosA)}$
= $\dfrac{sinA}{cosA}$
= $tanA$
= RHS
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Related Notes:
Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles
See all the solutions of Trigonometric Identities in this page.
Question: Prove that: (1 +sin2A -cos2A)\(1 +sin2A +cos2A) = tanA | SciPiPupil
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