If tanA = x/y, prove that y.cos2A +x.sin2A = y.
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:
Given,
tan A = $\frac{x}{y}$
Taking LHS;
= y * cos2A + x * sin2A
[ cos2A = $\dfrac{1-tan²A}{1+tan²A}$ and sin2A = $\dfrac{2tanA}{1+tan²A}$ ]
= $\dfrac{y *(1-tan²A)}{1+tan²A}$ +$\dfrac{x*2tanA}{1+tan²A}$
[ Replace tanA with (x/y) ]
= $\dfrac{y *\{1-\left (\dfrac{x}{y} \right)^2\}}{1+\left (\dfrac{x}{y} \right)^2}+ \dfrac{x * 2\dfrac{x}{y}}{1+\left (\dfrac{x}{y} \right)^2}$
= $\dfrac{\dfrac{y(y^2-x^2)}{y^2} + \dfrac{2x^2}{y}}{1+\left (\dfrac{x}{y}\right)^2}$
= $\dfrac{\dfrac{y^3-x^2y+2x^2y}{y^2}}{\dfrac{y^2+x^2}{y^2}}$
[ (a/b)/(c/d) = ad/bc. Since, b and d are same; we get: b/c ]
= $\dfrac{y(y^2+x^2)}{y^2+x^2}$
[ (y²+x²) in the numerator as well as denominator get cancelled ]
= y
= RHS
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Related Notes:
Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles
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