IIT JEE , AIEEE Forum - Mathematics , Trignometry

sathya_crazyteen

1. st 1+sin

-cos

/1+sin

+cos

+ 1+sin

+cos

/1+sin

-cos

=2 cosec

2...a=cos x+sin x;b=sin³x+cos³x;

find a³-3a+2b

3.if a sin² x+b cos ² x=c, b sin² x+ a cos²x=d and a tan x=b tan

(take x=

)

then st 1/a+1/b=1/c+1/d

Avinash_Bhat

1)

(1 + sin

- cos

) / (1 + sin

+cos

) + (1 + sin

+ cos

) / (1 + sin

+ cos

)

={ (1 + sin

- cos

)² + (1+ sin

+ cos

)² } / (1 +sin

+ cos

)(1 + sin

- cos

)

=(4 + 4sin

) / (2sin

+2sin²

)

=4(1 + sin

) / 2sin

(1 + sin

)

=2cosec

" Thathwamasi "

Avinash_Bhat

2)

a = (cos x + sin x),

b = (cos³x + sin³x) = (cos x + sin x)(1 - sin x cos x)

So, a³ - 3a + 2b = a(a² - 1) - (2a - b)

=(cos x +sin x)(2 cos x sin x) - (2 {cos x +sin x} - {cos x +sinx}{1 - sin x cosx} )

= " - (cos x + sin x)(1 + sin x cos x)

=(cos x + sin x) ( {2 cos x sin x} - {1 + sin x cos x} )

=(cos x + sin x)(cos x sin x - 1)

= - (cos³x + sin³x) = ( - b )

Please let me know if there are any mistakes...

" Thathwamasi "

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