Question

Sec A (1 – Sin A) (Sec A + Tan A)=……

(A) 0 (B) 1 (C) 2 (D) –1​

Answer 1

Answer:

A 0

Step-by-step explanation:

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Answer 2

Answer:

$\large\bf\underline\red{Question ➡} \\ \\ \sf \: SecA(1-SinA)(SecA+TanA) = \\ \\ \bf \: (A) =\sf \: 0 \\ \bf \: (B) = \sf \: 1 \sf \: \\ \bf \:( C ) =\sf \: 2 \\ \bf \: (D) =\sf - 1 \\ \\ \\ \large\bf\underline\red{answer \: ➡} \\ \\ \large{\boxed{\mathfrak\red{\fcolorbox{magenta}{aqua}{1}}}} \\ \\ \large\sf\underline\red{basic \: information \: ➡} \\ \\ ➥ \: \bf \: SecA = \sf \: \frac{1}{CosA} \\ \\ ➥ \: \bf \: TanA = \sf \: \frac{SinA}{CosA} \\ \\ ➥ \: \bf \: 1 - {SinA}^{2} = \sf \: {CosA}^{2} \\ \\ \\ \large\sf\underline\red{step \: of \: solution \: ➡} \\ \\ \sf \: SecA(1-SinA)(SecA+TanA) \\ \\ ➳ \: \sf \: \frac{1}{CosA} (1 - SinA) \bigg( \frac{1}{CosA} + \frac{SinA}{CosA} \bigg) \\ \\ \sf \: ➳ \: \frac{1}{CosA} (1 - SinA) \bigg( \frac{1 + SinA}{CosA } \bigg) \\ \\ \bf \: using \: formula \: ➴➴ \: \\ \sf (a + b) (a - b)= {(a)}^{2} - {(b)}^{2} \\ \\➳ \sf \: \frac{1}{CosA} \times \frac{1 - {SinA}^{2} }{CosA} \\ \\ ➳ \sf \: \frac{1 - {SinA}^{2} }{ {CosA}^{2} } \\ \\ \sf \: using \: \: \: ( 1 - {SinA}^{2} \: = {CosA}^{2} )\\ \\ ➳ \: \sf \: \frac{ {CosA}^{2} }{{CosA}^{2} } \\ \\ ➳ \: \sf \: \frac{ { \cancel{CosA}^{2}} }{{ \cancel{CosA}^{2}} } \\ \\ ➳ \: \: \large{\boxed{\mathfrak\red{\fcolorbox{magenta}{aqua}{1}}}}$

therefore option (B) is correct .

additional matter ➡

➥ cotA = 1/tanA

➥ cotA = cosA/sinA

➥ cosecA = 1/sinA

➥ sin²A + cos²A = 1

➳ 1 - sin²A = cos²A

➳ 1 - cos²A = sin²A

➥ 1 + tan²A = sec²A

➳ sec²A - tan²A = 1

➳ sec²A - 1 = tan²A

➥ 1 + cot²A = cosec²A

➳ cosec²A - 1 = cot²A

➳ cosec²A - cotA = 1