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Miscellaneous Exercise 3 (II) · Q90

Q.Prove the following: If sin⁡αsin⁡β−cos⁡αcos⁡β+1=0\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0 then prove cot⁡αtan⁡β=−1\cot\alpha\tan\beta=-1

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✓ Free question

Step 1: Given: sin⁡αsin⁡β−cos⁡αcos⁡β+1=0\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0, i.e. −(cos⁡αcos⁡β−sin⁡αsin⁡β)=−1-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-1, i.e. cos⁡(α+β)=1\cos(\alpha+\beta)=1.

Step 2: This forces α+β=0\alpha+\beta=0 (taking the principal solution), i.e. β=−α\beta=-\alpha.

Step 3: cot⁡αtan⁡β=cot⁡αtan⁡(−α)=cot⁡α⋅(−tan⁡α)=−(cot⁡αtan⁡α)=−1\cot\alpha\tan\beta=\cot\alpha\tan(-\alpha)=\cot\alpha\cdot(-\tan\alpha)=-(\cot\alpha\tan\alpha)=-1 (since cot⁡αtan⁡α=1\cot\alpha\tan\alpha=1).

✓Final answer

cot⁡αtan⁡β=−1\cot\alpha\tan\beta=-1

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