Question

Proof that ( sin4 theta - cos4 theta +1) cosec 2 theta =2

Answer 1

Step-by-step explanation:

In △ABC AB=AC

⇒∠B=∠C (Angles opposite to equal sides are equal)

Now using angle sum property

∠A+∠B+∠C=180

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⇒80

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+∠C+∠C=180

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⇒2∠C=180

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−80

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⇒∠C=

2

100

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=50

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now ∠C+∠x=180

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(Angles made on straight line (AC) are supplementary)

⇒50

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+∠x=180

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⇒∠x=180

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−50

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=130

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Answer 2

The equation $(sin^4\theta-cos^4\theta+1)cosec^2\theta=2$ is proved

Step-by-step explanation:

Given that $(sin^4\theta-cos^4\theta+1)cosec^2\theta=2$

To prove the equality :

That is to prove that LHS=RHS

Now taking LHS

$(sin^4\theta-cos^4\theta+1)cosec^2\theta$

• $=sin^4\theta (cosec^2\theta)-cos^4\theta (cosec^2\theta)+1(cosec^2\theta)$ ( applying the distributive property a(x+y+z)=ax+ay+az )
• $=sin^4\theta (\frac{1}{sin^2\theta})-cos^4\theta \frac{1}{sin^2\theta}+\frac{1}{sin^2\theta}$ ( by using the identity $cosec^2x=\frac{1}{sin^2x}$ )

• $=\frac{sin^4\theta-cos^4\theta+1}{sin^2\theta}$
• $=\frac{sin^4\theta+1-cos^4\theta}{sin^2\theta}$
• $=\frac{sin^4\theta+[(1^2)^2-(cos^2\theta)^2}{sin^2\theta}$
• $=\frac{sin^4\theta+[1^2-cos^2\theta][1^2+cos^2\theta]}{sin^2\theta}$ (by using the identity $a^2-b^2=(a+b)(a-b)$ )
• $=\frac{sin^4\theta+[1-cos^2\theta][1+cos^2\theta]}{sin^2\theta}$

• $=\frac{sin^4\theta+[sin^2\theta][1+cos^2\theta]}{sin^2\theta}$ ( here by using the identity $1-cos^2\theta=sin^2\theta$ )

• $=\frac{sin^4\theta}{sin^2\theta}+\frac{[sin^2\theta][1+cos^2\theta]}{sin^2\theta}$

• $=sin^2\theta+1+cos^2\theta$
• $=1+1$ ( by using the identity $sin^2x+cos^2x=1$ )
• $=2$ =RHS
• Hence we get that LHS=RHS

Hence the equation $(sin^4\theta-cos^4\theta+1)cosec^2\theta=2$ is proved