Question #04552

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Question #04552

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Answer 1 · 2017-05-12T04:20:55

Use $sin (θ) = cos (90^{\circ} - θ)$ $sin(theta)=cos(90^@-theta)$ . Answer: $2$ $2$

Explanation:

Evaluate ${(\frac{{sin 39}^{\circ}}{{cos 51}^{\circ}})}^{2} + {(\frac{{cos 51}^{\circ}}{{sin 39}^{\circ}})}^{2}$ $(sin39^@/cos51^@)^2+(cos51^@/sin39^@)^2$

Here, it is important to note that $sin (θ) = cos (90^{\circ} - θ)$ $sin(theta)=cos(90^@-theta)$ and $cos (θ) = sin (90^{\circ} - θ)$ $cos(theta)=sin(90^@-theta)$

Therefore, by substitution:

${(\frac{{sin 39}^{\circ}}{{cos 51}^{\circ}})}^{2} + {(\frac{{cos 51}^{\circ}}{{sin 39}^{\circ}})}^{2}$ $(sin39^@/cos51^@)^2+(cos51^@/sin39^@)^2$

$= {(\frac{cos (90^{\circ} - 39^{\circ})}{{cos 51}^{\circ}})}^{2} + {(\frac{{cos 51}^{\circ}}{cos (90^{\circ} - 39^{\circ})})}^{2}$ $=(cos(90^@-39^@)/cos51^@)^2+(cos51^@/cos(90^@-39^@))^2$

$= {(\frac{{cos 51}^{\circ}}{{cos 51}^{\circ}})}^{2} + {(\frac{{cos 51}^{\circ}}{{cos 51}^{\circ}})}^{2}$ $=(cos51^@/cos51^@)^2+(cos51^@/cos51^@)^2$

$= 1^{2} + 1^{2}$ $=1^2+1^2$

$= 2$ $=2$

Therefore, our answer is $2$ $2$ .

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Question #04552

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