Question

tan ∆ + sec 0∆ - 1 /tan∆-sec∆+1 =1+sin∆/cos∆ (∆=theta)​

Answer 1

Tan theta + sec theta -1/ tan - sec +1 = 1+ sincos

tan+sec -1+ tan-sec/tan-sec

tan²- sec² - 1 + tan - sec / tan - sec

tan + sec (tan - sec ) -1 + tan- sec) tan-sec

tan - sec( tan+sec-1+1)

/ tan- sec

tan + sec

.sin/cos + 1/ cos

1+ sin / cos