Question

(iv) tan (55° - A) - cot (35° + A)​

Answer 1

Answer:

$\huge\underline{\underline{\mathfrak\red{hola mate}}}$

$\huge\boxed{\fcolorbox{red}{lime}{Given}}$

tanA =cot(90-A).

therefore

tan(55-A)-cot(35+A)=tan55-tanA-cot(35+A)

=tan55-cot(90-A)-cot35+cotA....... from (1)=cot(90-55)-cot(90-A)-cot35+cotA.....since tanA =cot(90-A)

=cot35-cot(90-A)-cot35+cotA

= cotA-cot(90-A)

=0

$\huge\boxed{\fcolorbox{purple}{lime}{Thanks}}$

Answer 2

Answer:

ur answer:

tanA=cot(90-A)

therefore,

tan(55-A)-cot(35+A)

=tan55-tanA-cot(35+A)

tan 55-cot(90-A)-cot35+cotA

from(1)=cot(90-55)-cot(90-A)

cot35-cot(90-A)-cot35+cotA

=cot-cot(90-A)

=0

hope it helps u....

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thanku..

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