Question

sin square(-160)/sin square 70 + sin(180-theta)/sin theta= sec squared 20. How to prove this????​

Answer 1

Answer:

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Answer 2

$\frac{\sin^2(-160)}{\sin^270}+\frac{\sin(180-\theta)}{\sin \theta}= \sec^220$ Proved.

Step-by-step explanation:

Consider the provided information.

$\frac{\sin^2(-160)}{\sin^270}+\frac{\sin(180-\theta)}{\sin \theta}= \sec^220$

Consider the Left hand side.

Use the property: $\sin(-\theta)=\sin(\theta), \sin(180-\theta)=\sin\theta$

$\frac{\sin^2(-160)}{\sin^270}+\frac{\sin(180-\theta)}{\sin \theta}=\frac{\sin^2(160)}{\sin^270}+\frac{\sin(\theta)}{\sin \theta}$

$=\frac{\sin^2(180-20)}{\sin^2(90-20)}+1$

Use the property: $\sin(180-\theta)=\sin\theta, \sin(90-\theta)=\cos\theta$

$=\frac{\sin^2 20}{\cos^2 20}+1$

$=\tan^2 20+1$  (∴ sec²θ-tan²θ=1)

$=\sec^2 20$

LHS=RHS

Hence, proved

#Learn more

Prove that sec squared theta minus 1 into cot square theta is equal to 1​

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