Question

. मान ज्ञात कीजिए:
(i) sin 75°
X(ii) tan 15°​

Answer 1

Answer:

sin(75°) is numerically = sin(45+30)

Step-by-step explanation:

sin(a+b) = sinA•cosB+cosA•sinB

sin(45+30)= sin45•cos30+sin30•cos45

sin(45+30)= (1/√2)*(√3/2)+ (1/2)* (1/√2)

sin(90-15)= (√3/2√2) + (1/2√2)

sin75°=[(√3+1)/2√2]

Tan15°= Tan(45-30)

tan(A-B)= (tanA - tanB/1+tanA•tanB)

tan(45-30)= (tan45-tan30/1+tan45•tan30)

Tan(45-30)= (1-1/√3/ 1+ 1*1/√3) {Tan45=1, Tan30=1/√3}

Tan(45-30)=(√3-1/√3 /√3+ 1/√3)

tan15°= (√3-1/√3+1)

after rationalising we get,

(√3-1)^2/(√3-1)(√3+1)

3- 2√3+ 1/ 3-1

4- 2√3/ 2

(2-√3)

Tan15°= 2-√3

#sin75°=[(√3+1)/2√2]

#Tan15°= (2-√3)

hope it will be helpful....