Question

7. A triangle ABC is right angled at B; find the
sec A .sin C-tan Atan C
value of
sin B​

Answer 1

Step-by-step explanation:

A+90°+C=180°

=>A+C=90°

sinB

secAsinC−tanAtanC

=(secAsinC−

cosA

sinA

×

cosC

sinC

)÷sin90°

=(

cosA

sinC

−

cosA

sinA

×

cosC

sinC

)

=(

cosA

sin(90°−A)

−

cosA

sinA

×

cos(90°−A)

sin(90°−A)

)

=

cosA

cosA

−

cosA

sinA

×

sinA

cosA

=1−1

=0