Question

If sin(a+b)=1 and tan(a-b)=1/√3 then find tana + cotb and seca - cosecb

Answer 1

Answer:

$a + b =\pi \div 2$

$a - b = \pi \div 6$

Add on both steps we got 2a=

$\pi \div 3$

$a = \pi \div 3$

$b = \pi \div 6$

$tana + cotb = \sqrt{3} + \sqrt{3}$

$= 2 \sqrt{3}$

Seca-cosecb=0