Math, asked by shaili3, 1 year ago

please help me out in this question

Attachments:

shaili3: is it clear now
shaili3: ??

Answers

Answered by OOOIRKIOOO
1
HERE I M USING "A" INSTEAD OF "THEETA" SORRY..............

⇒ 1/sin(A) - sin(A) = l 

⇒ l² = 1/sin²(A) + sin²(A) - 2 

sec(A) - cos(A) = m 
⇒ 1/cos(A) - cos(A) = m 

⇒ m² = 1/cos²(A) + cos²(A) - 2 

l²m² = [1/sin²(A) + sin²(A) - 2] [1/cos²(A) + cos²(A) - 2]  

= 1/(sin²(A)cos²(A)) + cos²(A)/sin²(A) - 2/sin²(A) + sin²(A)/cos²(A) + sin²(A)cos²(A) - 2sin²(A) - 2/cos²(A) - 2cos²(A) + 4 

= 1/(sin²(A) + 1/cos²(A)) + (1 - sin²(A))/sin²(A) + (1 - cos²(A))/cos²(A) - 2(1/sin²(A) + 1/cos²(A)) - 2(sin²(A) + cos²(A)) + 4 + sin²(A)cos²(A)  

= -1/sin²(A) + 1/cos²(A) + 1/sin²(A) - 1 + 1/cos²(A) - 1 - 2 + 4 + sin²(A)cos²(A)  

= sin²(A)cos²(A) 

l² + m² + 3 = sin²(A) + 1/sin²(A) - 2 + cos²(A) + 1/cos²(A) - 2 + 3 

= 1 - 4 + 3 + 1/(sin²(A)cos²(A)) 

= 1/(sin²(A)cos²(A)) 

⇒ l²m² (l² + m² + 3) = sin²(A)cos²(A) / [sin²(A)cos²(A)] = 1

HOPE IT HELPS U SHAILI............
IF THEN MARK ME BRAINLIEST..................
@_ OOOIRKIOOO#
;P 

shaili3: Thank you very much Irki
shaili3: ...
shaili3: ...
shaili3: ...
shaili3: thanks
OOOIRKIOOO: ANYTYM SHAILI...........
Similar questions