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Question Number 159845 by abdullah_ff last updated on 21/Nov/21

$$\mathrm{if}q=1-sin\theta ;\mathrm{then}\mathrm{prove}\mathrm{that}$$$${(sec\theta -tan\theta )}^2=\frac{1}{q}$$

Commented by tounghoungko last updated on 21/Nov/21

$$\sin \theta =1-q$$$$\cos \theta =\sqrt{1-(1-2q+q^2)}=\sqrt{2q-q^2}$$$$\sec \theta =\frac{1}{\sqrt{2q-q^2}};\tan \theta =\frac{1-q}{\sqrt{2q-q^2}}$$$$\sec \theta -\tan \theta =\frac{q}{\sqrt{2q-q^2}}$$$${(\sec \theta -\tan \theta )}^2=\frac{q^2}{2q-q^2}$$

Answered by JDamian last updated on 21/Nov/21

$${\left(\frac{1-sin\theta }{cos\theta }\right)}^2=\frac{q^2}{{cos}^2\theta }=\frac{q^2}{1-{sin}^2\theta }=\frac{q^2}{1-{(1-q)}^2}=$$$$\frac{q^2}{1-1-q^2+2q}=\frac{q}{2-q}◼$$

Answered by som(math1967) last updated on 21/Nov/21

$${(sec\theta -tan\theta )}^2$$$$={\left(\frac{1-sin\theta }{cos\theta }\right)}^2$$$$=\frac{{(1-sin\theta )}^2}{{cos}^2\theta }$$$$=\frac{{(1-sin\theta )}^2}{(1+sin\theta )(1-sin\theta )}$$$$=\frac{1-sin\theta }{1+sin\theta }\ne \frac{1}{q}$$

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