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Please send the solution of attached question as soon as possible...... Properly plzzzzzzzzzz...........

Please send the solution of attached question as soon as possible......
Properly plzzzzzzzzzz...........

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Grade:11

1 Answers

mkmjnj
43 Points
5 years ago
Let's say   a= sin-1x   and    b = cos-1x
Due to lack of symbols let's denote pi by ¥
Also we know a+b = ¥/2
a3 -- b3 + (a-b)a.b = ¥3/16
(a-b)(a2 + b2 + a.b) + a.b(a-b)  =  ¥3/16
(a-b)[ a2 + b2 + 2a.b ]  = ¥3/16
(a-b)[a+b]2 = ¥3/16
(a-b)(¥2/4) = ¥3/16
(a-b) = ¥/4
¥/2 - 2b = ¥/4
2b = ¥/4
b = ¥/8
x = cos(¥/8) ....... ANS
 
 

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