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Question no.7 2nd part

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Hope u provide me right answer ☺✌⚡

Answer 1

$\sin( \alpha ) = 3 \div 5$
So,
$\tan( \alpha ) = 3 \div 4$
But it is mentioned that
$\alpha \: lies \: in \: 2nd \: quadrant \:$
Here sin is positive but tan is negative..
So,
$\tan( \alpha ) = - (3 \div 4)$
Similarly..
Since,
$\tan( \beta ) = 1 \div 2$
Hence,
$\sec( \beta ) = \sqrt{5}$
But it is mentioned that beta is in the IIIrd quad.. So, sec will be negative..
Therefore
$\sec( \beta ) = - \sqrt{5}$
Solve accordingly now.