#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Please give me a step wise solution of question with each & every reason

Faiz
107 Points
5 years ago
From 2nd eq in question...-> ax sinß / cos²ß = by cosß / sin²ß…-> tan³ß = by / ax-> tanß = (by/ax)⅓...From tanß you can take out sinß and cos ß which are respectively:::: sinß = (by)⅓ / sqroot of [ (by)⅔ + (ax)⅔ ]…cosß = (ax)⅓ / sqroot of [ (by)⅔ + (ax)⅔ ]…Putting this in 1st eq in the question....-> [ ax/ (ax)⅓ + by/ (by)⅓ ]*[sqroot of [ (by)⅔ + (ax)⅔ ] ] = a² - b².....-> [ (by)⅔ + (ax)⅔ ]⅔ = a² - b²....-> (by)⅔ + (ax)⅔ = ( a² - b ) ^ 3/2....Hence proved...
Faiz
107 Points
5 years ago
Correction for the last step::-> [ (by)⅔ + (ax)⅔ ] ^3/2( instead of 2/3 wrongly printed ) = a² - b²....-> (by)⅔ + (ax)⅔ = ( a² - b ) ^ 2/3......Hence proved...Sorry the powers were wrongly wriiten by me because of a little confusion but rest of the solution just before the last step remain s the same..