√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1

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$Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}$

$\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s$

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Which graph shows the solution to the system of linear inequalities?<br> y< 2x-5<br> y>-3x + 1

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Answer:

See explanation

Step-by-step explanation:

Plot the solution sets to both inequalities.

1. For the inequality $y$ First, plot the dotted line $y=2x+5$ (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

$2\cdot 0-5=-5$

so the origin does not belong to the needed part. Shade the part, which does not include origin.

2. For the inequality $y>-3x+1.$ First, plot the dotted line $y=-3x+1$ (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.