If `alpha+beta90^(@)` then the expression
`(tanalpha)/(tanbeta)+sin^(2)alpha+sin^(2)beta` is equal to:

Options

A. `tan^(2)alpha`
B. `tan^(2)beta`
C. `sin^(2)beta`
D. `sec^(2)beta`

Answer

d

Explanation

Given:
`alpha+beta=90^(@)`
to find `(tanalpha)/(tan beta)+sin^(@)alpha+sin^(2)beta=?`
`becausealpha+beta=90^(@)`
`rArralpha=90-beta`
`rArralpha,beta` are complementary angles
`rArr(tan alpha)/(tan(90-alpha))+sin^(2)alpha+sin^(2)(90-alpha)`
`rArr(tanalpha)/(cotalpha)+sin^(2)alpha+cos^(2)alpha`
`rArrtan^(2)alpha+1`
`(because sin^(2)alpha+cos^(2)alpha=1)`
`rArrsec^(2)alpha(because 1+tan^(2)alpha=sec^(2)alpha)`