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cluponka [151]
4 years ago
13

Solve for y . =6y2 Simplify your answer as much as possible.

Mathematics
1 answer:
Scorpion4ik [409]4 years ago
3 0
Y could be twelve, but I think I'm a 100% wrong. :'D
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Answer:

Let \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right), we proceed to prove the trigonometric expression by trigonometric identity:

1) \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right) Given

2) \left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)   \tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}

3) \left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)    

4) \left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)    \sin^{2}A+\cos^{2}A = 1

5) \frac{1}{\sin^{2}A\cdot \cos^{2}A}

6) \frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}    \sin^{2}A+\cos^{2}A = 1

7) \frac{1}{\sin^{2}A-\sin^{4}A} Result

Step-by-step explanation:

Let \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right), we proceed to prove the trigonometric expression by trigonometric identity:

1) \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right) Given

2) \left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)   \tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}

3) \left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)    

4) \left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)    \sin^{2}A+\cos^{2}A = 1

5) \frac{1}{\sin^{2}A\cdot \cos^{2}A}

6) \frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}    \sin^{2}A+\cos^{2}A = 1

7) \frac{1}{\sin^{2}A-\sin^{4}A} Result

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