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1-SinQ = (sec Q tanQ)2
1+sinQ​

Sagot :

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Answer:

The answer is below

Explanation:

Prove that:

[tex]\frac{1+sinQ}{1-sinQ}=(secQ + tanQ)^2[/tex]

Trigonometric identities are equalities involving trigonometric functions for which both sides of the equality are equal and defined. Some trigonometric identities are:

sin²Ф + cos²Ф = 1;  1/cosФ = secФ; 1/sinФ = cosecФ; cosФ/sinФ = cotФ; 1 +  tan²Ф = sec²Ф

Given:

[tex]\frac{1+sinQ}{1-sinQ}\\\\Divide\ through\ by \ cosQ:\\\\ \frac{\frac{1}{cosQ} +\frac{sinQ}{cosQ} }{\frac{1}{cosQ} -\frac{sinQ}{cosQ} }=\frac{secQ+tanQ}{secQ-tanQ}\\\\Next, rationalize\ the\ denominator\ by \ multiplying\ the\ numerator \ and\ \\denominator\ by\ secQ+tanQ:\\\\\frac{secQ+tanQ}{secQ-tanQ}*\frac{secQ+tanQ}{secQ+tanQ}=\frac{(secQ+tanQ)^2}{sec^2Q+secQtanQ-secQtanQ-tan^2Q}\\\\=\frac{(secQ+tanQ)^2}{sec^2Q-tan^2Q} ;\ But sec^2Q-tan^2Q=1,hence:\\\\[/tex]

[tex]\frac{(secQ+tanQ)^2}{sec^2Q-tan^2Q} =\frac{(secQ+tanQ)^2}{1}=(secQ+tanQ)^2\\\\\frac{1+sinQ}{1-sinQ}=(secQ+tanQ)^2[/tex]

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