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In a separate post from you yesterday, I tried to substantiate your query. You can go through.
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9 years agoBeginner-Level 2
tan13A-tan9A tan4A=tan13A tan9A tan4A
For this equation, we will use the formula:
I) tan 2a= tan(a + a) and,
II) tan (a + b)= (tan a+ tan b)/(1-tan a tan b)
From (I), we can say
tan 13A= tan (9A+4A)
but, tan(9A+4A)=(tan 9A + tan 4A)/ (1-tan 9Atan 4A)
Therefore tan13A=(tan 9A+tan4A)/(1-tan9Atan4A)
=tan13A-tan13Atan9Atan4A=tan 9A+tan 4A
Therefore,
tan13A-tan9A-tan4A=tan13A tan9A tan4A
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