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
Menu
Aditya Sharma Grade: 12
        

\sqrt{a-b/a+b}=tanAtanB then show that 
(a-bCos2A)(a-bCos2B)=a^{2}-b^{2}

5 years ago

Answers : (1)

u. s. p. srinivas aditya
32 Points
										

sqrt{a-b/a+b}=tanAtanB  (given)

multiplying and dividing (a-b) in LHS...

sqrt{(a-b)(a-b)/(a+b)(a-b)}=tanAtanB

sqrt{(a-b)^2/a^2-b^2}=tanAtanB

squaring both sides...

(a-b)^2/a^2-b^2=tan^2Atan^2B

a^2-b^2=(a-b)^2/tan^2Atan^2B   ~eq.[1]

 

To prove: (a-bCos2A)(a-bCos2B)=a^2-b^2

from eq.[1]

we can prove,  (a-b)^2/tan^2Atan^2B=(a-bCos2A)(a-bCos2B)

RHS-

=> {a-b[(1-tan^2A)/(1+tan^2A)]}{a-b[(1-tan^2B)/(1+tan^2B)]} 

=> [a-(b-btan^2A)/(1+tan^2A)][a-(b-btan^2B)/(1+tan^2B)]

=> [(a+atan^2A-b+btan^2A)(a+atan^2B-b+btan^2B)]/[1+tan^2A+tan^2B+tan^2Atan^2B]

on solving further....

=> (a^2+b^2-2ab)(tan^2B)/(tan^2Atan^4B)

=> (a-b)^2/tan^2Atan^2B

                          hence proved...!! 

5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details