If cos A=a÷b prove that sinA×tanA÷secA-1=a+b÷b
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Step-by-step explanation:
The trigonometric ratios which will be used in this question are: secA = 1/cosA, tanA = sinA/cosA
The identity used will be (a - b)(a + b) = (a2 - b2)
Taking the left hand side,
secA(1 - sinA)(secA + tanA)
= (1/cosA)(1 - sinA)(1/cosA + sinA/cosA)
= (1/cosA)(1 - sinA)(1+ sinA)/cosA
= (1/cosA)(1/cosA)(1 - sin2A)
= (1/cos2A)(cos2A) Since 1 - sin2A = cos2A
= 1
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