Question

If cosec theta-sin theta=l; sec theta-cos theta=m then prove that l^m^(l^+m^+3)=1

Answer 1

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The solution is in the pic..

Answer 2

Solution:

As the question says $\csc \theta-\sin \theta=l \text { and } \sec \theta-\cos \theta=m$

Let us take$\csc \theta-\sin \theta=l$

$\frac{1}{\sin \theta}-\sin \theta=l$

$\frac{1-\sin ^{2} \theta}{\sin \theta}=l$

$\frac{\cos ^{2} \theta}{\sin ^{2} \theta}=l$

Let us now simplify the m, that is $\sec \theta-\cos \theta=m$

After simplifying we get $m=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$

Now putting the values of m and n in $l^{2} m^{2}\left(l^{2}+m^{2}+3\right)=1$we get

$\frac{\cos ^{2} \theta^{2}}{\sin ^{2} \theta} \cdot \frac{\sin ^{2} \theta^{2}}{\cos ^{2} \theta} \cdot\left(\frac{\cos ^{2} \theta^{2}}{\sin ^{2} \theta}+\frac{\sin ^{2} \theta^{2}}{\cos ^{2} \theta}+3\right)=1$

Simplifying the equation, we get

$\left(\cos ^{2} \theta\right)^{3}+\left(\sin ^{2} \theta\right)^{3}+3 \sin ^{2} \theta \cos ^{2} \theta=\left(\cos ^{2} \theta+\sin ^{2} \theta\right)^{3}$

Therefore, as we can see that the value of $l^{2} m^{2}\left(l^{2}+m^{2}+3=1\right$. Hence RHS = LHS.

Hence, Proved.