Question

prove that sin 70°-sin 15°÷cos 15°- cos 75°=1​

Answer 1

Answer:

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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