Question

please prove this:
(1+sec2A *cot2Y)/(1+sec2B *cot2 Y )=(1+tan2A*cos2Y)/(1+tan2B*cos2Y)

in above 2 is power in all case

Answer 1

Answer on question #34318 – Math – Trigonometry

Please prove this:

$$(1 + \sec 2 A * \cot 2 Y) / (1 + \sec 2 B * \cot 2 Y) = (1 + \tan 2 A * \cos 2 Y) / (1 + \tan 2 B * \cos 2 Y)$$

In above 2 is power in all case

Proving

$$\frac {1 + \sec^ {2} A \cot^ {2} Y}{1 + \sec^ {2} B \cot^ {2} Y} = \frac {1 + \frac {1}{\cos^ {2} A} \frac {\cos^ {2} Y}{\sin^ {2} Y}}{1 + \frac {1}{\cos^ {2} B} \frac {\cos^ {2} Y}{\sin^ {2} Y}} = \frac {\cos^ {2} B \sin^ {2} Y (\cos^ {2} A \sin^ {2} Y + \cos^ {2} Y)}{\cos^ {2} A \sin^ {2} Y (\cos^ {2} B \sin^ {2} Y + \cos^ {2} Y)} =$$ $$= \frac {\cos^ {2} B (\cos^ {2} A \sin^ {2} Y + 1 - \sin^ {2} Y)}{\cos^ {2} A (\cos^ {2} B (1 - \cos^ {2} Y) + \cos^ {2} Y)} = \frac {\cos^ {2} B (1 - \sin^ {2} Y (1 - \cos^ {2} A))}{\cos^ {2} A (\cos^ {2} B - \cos^ {2} B \cos^ {2} Y + \cos^ {2} Y)} =$$ $$= \frac {\cos^ {2} B (\sin^ {2} A + \cos^ {2} A - \sin^ {2} Y \sin^ {2} A)}{\cos^ {2} A (\cos^ {2} B + \cos^ {2} Y (1 - \cos^ {2} B))} = \frac {\cos^ {2} B (\cos^ {2} A + \cos^ {2} Y \sin^ {2} A)}{\cos^ {2} A (\cos^ {2} B + \cos^ {2} Y \sin^ {2} B)} =$$ $$= \frac {\frac {(\cos^ {2} A + \cos^ {2} Y \sin^ {2} A)}{\cos^ {2} A}}{\frac {(\cos^ {2} B + \cos^ {2} Y \sin^ {2} B)}{\cos^ {2} B}} = \frac {(1 + \tan^ {2} A \cos^ {2} Y)}{(1 + \tan^ {2} B \cos^ {2} Y)}.$$

QED.

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