Question

TRIGO...PLEASE....my friends

Answer 1

$\huge\boxed{\texttt{\fcolorbox{red}{yellow}{Heya Mate!!!}}}$

HERE IS THE SOLUTION OF YOUR QUESTION⤵⤵⤵⤵⤵⤵


SOLUTION:-

GIVEN:-

SECØ = 13/5

PROVE THAT:- (2COSØ - 3COSØ)/(4SINØ - 9COSØ) = 3

PROOF:-

WE HAVE ,

SECØ = 13/5

THEN , COSØ = 5/13 [COS IS THE RECIPROCAL OF SEC]


SIN²Ø + COS²Ø = 1

SIN²Ø = 1 - COS²Ø

SIN²Ø = 1 - (5/13)²

SIN²Ø = 1 - (25/169)

SIN²Ø = (169 - 25)/169

SIN²Ø = 144/169

(SINØ)² = (12/13)²

SINØ = 12/13

PUT THESE VALUES IN ...EQN

(2COSØ - 3COSØ)/4SINØ - 9COSØ) = 3

LHS

= (2 × 5/13 - 3 × 5/13) / (4 × 12/13 - 9 × 5/13)

= 5/13(2 - 3) / [( 48/13) - (45/13)]

= 5/13(-1) / [ (48 - 45)/13 ]

= (-5/13 ) / ( 3/13)

= -5/3

LHS != RHS...


HENCE PROVED...


HOPE YOU LIKE IT....

#KRISHN12❤❤

Answer 2

hey frnd hope this will help uh....
nd plzz mark as brainliest if it helps uh ..