Question

If cos B = 15/17 find sin B tan B cot B explain in detail plz​

Answer 1

Answer:

sinA=15/17 and cosB=12/13

Using, sin²A+cos²A=1 we get,

cos²A=1-sin²A=1-225/289=(289-225)/289=64/289

cosA=8/17 (neglecting the neggative sign since

A,B are positive acute angles)

sin²B=1-cos²B=1-144/169=(169-144)/169=25/169

sinB=5/13 (neglecting the neggative sign since

A,B are positive acute angles)

sin(A+B)

=sinAcosB+cosAsinB

=15/17×12/13+8/17×5/13

=180/221+40/221

=220/221

cos(A-B)

=cosAcosB-sinAsinB

=8/17×12/13-15/17×5/13

=96/221-75/221

=21/221

tanA=sinA/cosA=(15/17)/(8/17)=15/8 and

tanB=sinB/cosB=(5/13)/(12/13)=5/12

tan(A+B)

=(tanA+tanB)/(1-tanAtanB)

=(15/8+5/12)/(1-15/8×5/12)

=[(45+10)/24]/[(96-75)/96]

=(55/24)/(21/96)

= 55/24×96/21

=210/21

=10

Step-by-step explanation:

@GENIUS