Question

if 1 + sin^2 alpha = 3 sin alpha.cos alpha.. then find cot alpha​

Answer 1

Here we have

$1+sin^{2} \alpha =3sin\alpha .cos\alpha$

Now put the value

$sin^{2} \alpha +cos^{2} \alpha =1$

So we get

$sin^{2} \alpha +cos^{2} \alpha +sin^{2} \alpha =3sin\alpha cos\alpha \\2sin^{2} \alpha +cos^{2} \alpha =3sin\alpha cos\alpha \\2sin^{2} \alpha +cos^{2} \alpha -3sin\alpha cos\alpha =0\\2sin^{2} \alpha -3sin\alpha cos\alpha +cos^{2} \alpha=0\\2sin^{2} \alpha -2sin\alpha cos\alpha-sin\alpha cos\alpha +cos^{2}\alpha =0 \\2sin \alpha (sin\alpha- cos\alpha) -cos\alpha (sin\alpha-cos\alpha )=0$

So after solving above trigonometric terms we get

$(2sin\alpha -cos\alpha )-(sin\alpha -cos\alpha )=0$

Now first we solve

$2sin\alpha -cos\alpha =0\\cos\alpha =2sin\alpha \\cos\alpha /sin\alpha =2\\cot\alpha =2$

Similarly we can solve

$sin\alpha -cos\alpha =0\\sin\alpha =cos\alpha \\cos\alpha /sin\alpha =1\\cot\alpha =1$

Hence, we have find that

$cot\alpha =1,2$

Answer 2

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