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Mukul Dinesh Om
Asked: 2022-11-10T19:22:23+05:30 In:

If `alpha+beta90^(@)` then the expression
`(tanalpha)/(tanbeta)+sin^(2)alpha+sin^(2)beta` is equal to:
A. `tan^(2)alpha`
B. `tan^(2)beta`
C. `sin^(2)beta`
D. `sec^(2)beta`

If `alpha+beta90^(@)` then the expression
`(tanalpha)/(tanbeta)+sin^(2)alpha+sin^(2)beta` is equal to:
A. `tan^(2)alpha`
B. `tan^(2)beta`
C. `sin^(2)beta`
D. `sec^(2)beta`
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1 Answer

  1. Correct Answer – d
    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)`

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