Question

$sec ^{2} 48 + \sec^{2} 42$
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Answer 1

Given;

• sec²48 + sec²42
• sec²48 + sec²(90-48). ( sec(90-A) = cosecA )
• sec²48 + cosec²48.
• 1/cos²48 + 1/sin²48. (secA=1/cosA). (cosecA= 1/sinA)
• ( sin²48 + cos²48 ) / cos²48×sin²48
• 1/cos²48×sin²48
• sec²48×cosec²48

#answerwithquality

#BAL