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Question

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

1
$- 1$
0
None of these

The correct answer is: 1

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 rightwards double arrow alpha plus beta equals fraction numerator pi over denominator 2 end fraction sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction rightwards double arrow alpha minus beta equals fraction numerator pi over denominator 6 end fraction$
Solving, we get $alpha equals pi divided by 3 blank a n d blank beta equals pi divided by 6$
Now $blank tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses equals tan invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus fraction numerator pi over denominator 3 end fraction close parentheses tan invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus fraction numerator pi over denominator 6 end fraction close parentheses equals tan invisible function application fraction numerator 2 pi over denominator 3 end fraction tan invisible function application fraction numerator 5 pi over denominator 6 end fraction$
$equals open parentheses negative cot invisible function application fraction numerator pi over denominator 3 end fraction close parentheses open parentheses negative cot invisible function application fraction numerator pi over denominator 6 end fraction close parentheses equals open parentheses negative fraction numerator 1 over denominator square root of 3 end fraction close parentheses open parentheses negative square root of 3 close parentheses equals 1$

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