1) If tanA = cot (2A – 39°), where A is an acute angle in right angles triangle, 
   find the value of A.

2) If tan A = $1/√3$ , find cosec A.

3) Prove that 

(i) $√({1+cosA}/{1-cosA})$ = $cosecA - cotA$

(ii) $√({1+cosA}/{1-cosA})$ = ${(1-cosA)/{sinA}}$

(iii) ${(1+sinA)(1-sinA)}/{(1+cosA)(1-cosA)}$ = $(cos^2A - sin^2A)$ = $cot^2A$

(iv) ${1-tan^2A}/{1+tan^2A}$ = ${cot^2A-1}/{cosec^2A}$

(v) $tanA + cotA+ 1/{tanA} + 1/{cotA} = 2cosecA secA$

(vi)  $(sin^4A+cos^4A)/{1-2sin^2Acos^2A}$ = $1$

(vii) $(sinA+cosA)^3/{1+2sinAcosA}$ = ${secA+cosecA}/{cosecAsecA}$


Solution