Question

find the value of i) cos 105°ii) sin105° iii) tan 7π/12 plss tell me the answer​

Answer 1

We have

LHS = sin105° + cos105°

= sin(60 + 45 ) + cos ( 60 + 45 )

= ( sin 60 cos 45 + cos 60 sin 45 )

+ ( cos 60 cos 45 - sin60 din 45)

= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)

= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2 = RHS

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Answer 2

Cos 105
= cos (45+60)
=> Cos(a+b)= cosA cosB - sinA sinB
It becomes
Cos45 x cos60 - sin45 x sin60
= 1/root2 x 1/2 - 1/root2 x root3/2
= 1/ 2root2 - root3/2root2
= 1 - root3 / 2root2

Sin 105
= sin(45+60)
=> sin(A+B) = sinA cosB + cosA sinB
So it becomes
Sin45 x cos60 + cos45 x sin60
= 1/root2 x 1/2 + 1/root2 x root3/2
= 1/2root2 +. Root3/ 2root2
= 1 + root3 / 2root2

Tan 7 pie/12
7pie/12 = 7pie/12 x 180/pie = 75 degree
So, it becomes
Tan (75) = tan (45+30)
=> tan(A+B) = tanA + TanB / 1 - tanA tanB
So it becomes
Tan 45 + tan30 / 1 - tan 45 x tan 30
= 1 + 1/root3 / 1- 1/root3
= root3 +1 / root3 - 1