If sinA=1/2 find valueof 2secA/1+tan*(square)A
Answers
Answered by
317
Hi ,
sinA = 1/2
sinA = sin30°
Therefore ,
A = 30°
Now ,
2secA / ( 1 + tan² A )
= 2secA / sec²A ( since 1+ tan² A = sec² A )
= 2 / secA
= 2 cos A
= 2 cos30°
= 2 × √3 /2
= √3
I hope this helps you.
:)
sinA = 1/2
sinA = sin30°
Therefore ,
A = 30°
Now ,
2secA / ( 1 + tan² A )
= 2secA / sec²A ( since 1+ tan² A = sec² A )
= 2 / secA
= 2 cos A
= 2 cos30°
= 2 × √3 /2
= √3
I hope this helps you.
:)
Answered by
30
Answer:
May this help u ...............
Attachments:
Similar questions