Question

If sin alpha = 1/3, evaluate tan alpha×sec alpha+cos alpha×cosec alpha.​

Answer 1

Answer:

I have taken alpha as A here

so ...sinA=1/3

cosA=√1-(1/3)^2=√1-1/9=2√2/3

now. tanA×secA+cosA×cosecA

=1/2√2 × 3/2√2 + 2√2/3 × 3/1

=3/8 + 2√2/1

=(3+16√2)/8

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Answer 2

SIN alpha = 1/3

using Pythagoras theorem

P = 1

B = √8

H = 3

putting values in equation

= 1/√8*3/√8+√8/3*3

= 3/8+√8. [√8 = 2]

ans = (3+8√8)/8 or (3+16√2)/8