Question 134

The value of $$\frac{sin 43^{\circ}}{cos 47^{\circ}}+\frac{cos 19^{\circ}}{sin 71^{\circ}}-8cos^{2}60^{\circ}$$ is

Solution

we need to find value of $$\frac{sin 43^{\circ}}{cos 47^{\circ}}+\frac{cos 19^{\circ}}{sin 71^{\circ}}-8cos^{2}60^{\circ}$$

$$\frac{sin (90-47)}{cos 47}+\frac{cos (90-17)}{sin 71}-8cos^{2}60$$

$$\frac{cos47)}{cos 47}+\frac{sin17)}{sin 71}-8( \frac{1}{2})^2$$

= 0

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