Question

if sin(A+B)=1 and tan(A-B)=1/root 3,find the value of 1)tanA+cotA 2)secA-cosecB

Answer 1

sec(A+B)=1                [∵ sin 90°=1 ]

 ⇒(A+B)=90°       ............(1)


tan(A-B)=1/√3          [∵ tan 30°=1/√3 ]

⇒(A-B)=30°        .............(2)

 Solving (1) and (2),we get,

A=60°
B=30°


1) tan A + cot B

tan 60°+ cot 30°
= √3 + √3
= 2√3


2) sec A - cosec B

sec 60° - cosec 30°
= 2-2
= 0

Answer 2

Answer:

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