2 years ago

If 2tanB+cotB=tanA,then proved that 2tan(A-B)=cotB

Mathematics

1 answer:

Stolb23 [73]2 years ago

6 0

Answer:

it is so hard it's optional math bro I tried much

Step-by-step explanation:

hope others can give answers

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Please Hurry ...Which expression is equivalent to

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Answer:

$\huge\boxed{\sf \frac{160rs^5}{t^6}}$

Step-by-step explanation:

$\sf 5r^6t^4 ( \frac{4r^3s^tt^4}{2r^4st^6} ) ^5$

Using rule of exponents $\sf a^m/a^n = a^{m-n}$

$\sf 5r^6t^4 ( 2 r^{3-4} s^{2-1}t^{4-6})^5\\5r^6t^4(2r^{-1}st^{-2})^5\\5r^6t^4 * 32 r^{-5}s^5t^{-10}$

Using rule of exponents $\sf a^m*a^n = a^{m+n}$

$\sf 160 r^{6-5}s^5t^{4-10}$

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To equalize the negative sign, we'll move t to the denominator

$\sf \frac{160rs^5}{t^6}$

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If 2tanB+cotB=tanA,then proved that 2tan(A-B)=cotB​

If 2tanB+cotB=tanA,then proved that 2tan(A-B)=cotB