`[[1,-tan alpha],[tan alpha,1]][[1,tan alpha],[-tan alpha,1]]=[[a,-b],[-b,a]]` then `sec "^(2)alpha`

1+tan alpha*tan((alpha)/(2))

[[1,-tan((alpha)/(2))tan((alpha)/(2)),1]][[1,-tan((alpha)/(2))tan((alpha)/(2)),1]]

(1)/(tan3 alpha-tan alpha)-(1)/(cot3 alpha-cot alpha)=cot2 alpha

(1+tan alpha tan beta)^(2)+(tan alpha-tan beta)^(2)=

(1-2sin^(2)alpha)/(1+sin2 alpha)=(1-tan alpha)/(1+tan alpha)

(1-2sin^(2)alpha)/(1+sin2 alpha)=(1-tan alpha)/(1+tan alpha)

P=[[sec alpha,tan alpha-cot alpha,cos alpha]] and Q=[[-cos alpha,tan alpha-cot alpha,-sec alpha]]

If sec alpha=(5)/(4), evaluate (1-tan alpha)/(1+tan alpha)

Prove the following identity: (1+tan alpha tan beta)^(2)+(tan alpha)(tan alpha-tan beta)^(2)=sec^(2)alpha sec^(2)beta