Question

if tanA = 1/7 and sinB = 1/√10, then find tan(A+2B)​

Answer 1

Answer:

here's the solution in the attachment above

Answer 2

Answer:

1 ans.

Step-by-step explanation:

we have, tanA = 1/7

sinB = 1/√10

to find, tan(A+2B)=[(tanA+tan2B)/

(1-tanAtan2B)]

as, tan2B = 2tanB/(1-tan²B)

so as, sinB = 1/√10

cosB = √{1-1/10}

= √{9/10}

= 3/√10

tanB = (1/√10)/(3/√10)

= 1/3

so, tan2B = (2×1/3)/(1-1/9)

= (2/3)/(8/9)

= 3/4

so, tan(A+2B) = [(1/7+3/4)/(1-1/7×3/4)]

= [(25/28)/(25/28)]

= 1 ....ans.