prove that sin 70°-sin 15°÷cos 15°- cos 75°=1
Answers
Step-by-step explanation:
Answer:
\frac{sin75-sin15}{cos75+cos15}
cos75+cos15
sin75−sin15
=\frac{\sqrt{3}}{3}
3
3
Explanation:
Given \frac{sin75-sin15}{cos75+cos15}
cos75+cos15
sin75−sin15
=\frac{sin(90-15)-sin15}{cos(90-15)+cos15}
cos(90−15)+cos15
sin(90−15)−sin15
/* i) sin(90-A) = cosA
ii)cos(90-15) = sin15 */
=\frac{cos15-sin15}{sin15+cos15}
sin15+cos15
cos15−sin15
Multiply numerator and denominator by (cos15 - sin15), we get
= \frac{(cos15-sin15)(cos15-sin15)}{(cos15+sin15)(cos15-sin15)}
(cos15+sin15)(cos15−sin15)
(cos15−sin15)(cos15−sin15)
= \frac{(cos15-sin15)^{2}}{cos^{2}15-sin^{2}15}
cos
2
15−sin
2
15
(cos15−sin15)
2
/* x²-y² = (x+y)(x-y) */
= \frac{cos^{2}15+sin^{2}15-2cos15sin15}{cos(2\times15)}
cos(2×15)
cos
2
15+sin
2
15−2cos15sin15
/* cos²A - sin²A = sin2A */
=\frac{1-sin(2\times15)}{cos30}
cos30
1−sin(2×15)
/* 2sinAcosA = sin2A */
= \frac{1-sin30}{cos30}
cos30
1−sin30
= \frac{1-\frac{1}{2}}{\frac{\sqrt{3}}{2}}
2
3
1−
2
1
= \frac{\frac{(2-1)}{2}}{\frac{\sqrt{3}}{2}}
2
3
2
(2−1)
= \frac{1}{\sqrt{3}}
3
1
Rationalising the denominator, we get
= \frac{\sqrt{3}}{\sqrt{3}\times \sqrt{3}}
3
×
3
3
= \frac{\sqrt{3}}{3}
3
3