Question

(Class 10 cbse-maths)
If cosec theta-sin theta =a^3 and sec theta -cos theta = b^3, prove that a^2b^2 (a² +b²)=1

Answer 1

$\textbf{Given:}$

$a^3=cosec\theta-sin\theta$

$a^3=\displaystyle\frac{1}{sin\theta}-sin\theta$

$a^3=\displaystyle\frac{1-sin^2\theta}{sin\theta}$

$a^3=\displaystyle\frac{ cos^2\theta}{sin\theta}$

$a=\displaystyle\frac{ cos^\frac{2}{3}\theta}{sin^\frac{1}{3}\theta}$

$\text{similarly,}$

$b^3=sec\theta-cos\theta$

$\implies\,b=\displaystyle\frac{ sin^\frac{2}{3}\theta}{cos^\frac{1}{3}\theta}$

$\text{Now,}$

$a^2b^2(a^2+b^2)$

$=\displaystyle(\frac{ cos^\frac{4}{3}\theta}{sin^\frac{2}{3}\theta})(\frac{ sin^\frac{4}{3}\theta}{cos^\frac{2}{3}\theta})(\frac{ cos^\frac{4}{3}\theta}{sin^\frac{2}{3}\theta}+\frac{ sin^\frac{4}{3}\theta}{cos^\frac{2}{3}\theta})$

$=\displaystyle(\frac{ cos^\frac{4}{3}\theta}{sin^\frac{2}{3}\theta})(\frac{ sin^\frac{4}{3}\theta}{cos^\frac{2}{3}\theta})(\frac{ cos^2\theta+sin^2\theta}{sin^\frac{2}{3}\theta\,cos^\frac{2}{3}\theta})$

$=\displaystyle(\frac{ cos^\frac{4}{3}\theta}{sin^\frac{2}{3}\theta})(\frac{ sin^\frac{4}{3}\theta}{cos^\frac{2}{3}\theta})(\frac{1}{sin^\frac{2}{3}\theta\,cos^\frac{2}{3}\theta})$

$=\displaystyle\,cos^\frac{4}{3}\theta\,sin^\frac{4}{3}\theta(\frac{1}{sin^\frac{4}{3}\theta\,cos^\frac{4}{3}\theta})$

$=1$

$\therefore\boxed{\bf\,a^2b^2(a^2+b^2)=1}$

Find more:

X/acosθ+y/bsinθ=1 and x/asinθ-y/bcosθ=1, prove that x²/a²+y²/b²=2.

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Answer 2

Answer:

ANSWER IN GIVEN LINK:-

If x= cosecA - sinA and y= secA - cosA, then show that x^2/3 + y^2/3 = ( xy)^-2/3

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