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2 years ago

Prove that : 1/1-secA + 1/1+secA = -2cot^2A

Mathematics

1 answer:

irina1246 [14]2 years ago

7 0

Answer:

Step-by-step explanation:

LHS = 1/1-secA + 1/1+secA

$= \frac{1*(1+Sec A)}{(1-Sec A)(1+Sec A)}+\frac{1*(1-Sec A)}{(1+Sec A)*(1-Sec A)}\\\\=\frac{(1+Sec A)}{1-Sec^{2} A}+\frac{(1-Sec A)}{1-Sec^{2} A}\\\\=\frac{1+Sec A+1-Sec A}{(1-Sec^{2} A)}\\\\= \frac{2}{(1-Sec^{2} A)}\\\\=\frac{2}{-(Sec^{2} A - 1)}\\\\= \frac{-2}{Tan^{2} A}\\\\= -2Cot^{2} A$

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2 years ago

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sweet [91]

Answer:

Equation 1 = Equation 2

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Step-by-step explanation:

Let x represent the number of 30-day periods.

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Therefore the equation =

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= 165 - 4x.......... Equation 1

He has another gift card for $145 that loses $3.50 for each 30-day period it is not used.

$145 -$3.50 × x

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2 years ago

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Answer:

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2 years ago

Prove that : 1/1-secA + 1/1+secA = -2cot^2A​

Prove that : 1/1-secA + 1/1+secA = -2cot^2A