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Question Number 155810 by zainaltanjung last updated on 05/Oct/21

$$\mathrm{Verify}\mathrm{the}\mathrm{identity}\mathrm{in}\mathrm{Excercise}\mathrm{below}$$$$1).\cos \theta \sec \theta =1$$$$2).(1+\cos \beta )(1-\cos \beta )=\sin {}^2\beta$$$$3).\cos {}^2\mathrm{x}(\sec {}^2\mathrm{x}-1)=\sin {}^2\mathrm{x}$$$$4).\frac{\sin \mathrm{t}}{\mathrm{cosec}\mathrm{t}}+\frac{\cos \mathrm{t}}{\sec \mathrm{t}}=1$$$$5).\frac{\mathrm{cosec}{}^2\theta }{1+\tan {}^2\theta }=\cot {}^2\theta$$

Answered by puissant last updated on 05/Oct/21

$$1)$$$$cos\theta sec\theta =cos\theta \times \frac{1}{cos\theta }=1..$$$$2)$$$$(1-cos\beta )(1+cos\beta )=1-{cos}^2\beta ={sin}^2\beta ..$$$$3)$$$${cos}^{2}x({sec}^{2}x-1)={cos}^{2}x(\frac{1}{{cos}^{2}x}-1)$$$$=1-{cos}^{2}x={sin}^{2}x..$$$$4)$$$$\frac{sint}{cosect}+\frac{cost}{sect}=\frac{sint}{\frac{1}{sint}}+\frac{cost}{\frac{1}{cost}}={sin}^{2}t+{cos}^{2}t=1..$$$$5)$$$$\frac{{cosec}^2\theta }{1+{tan}^2\theta }=\frac{1}{{sin}^2\theta }\times \frac{1}{1+\frac{{sin}^2\theta }{{cos}^2\theta }}=\frac{1}{{sin}^2\theta }\times \frac{{cos}^2\theta }{{cos}^2\theta +{sin}^2\theta }$$$$=\frac{1}{{sin}^2\theta }\times \frac{{cos}^2\theta }{1}={\left(\frac{cos\theta }{sin\theta }\right)}^2={cot}^2\theta ..$$

Commented by zainaltanjung last updated on 05/Oct/21

$$\mathrm{All}\mathrm{of}\mathrm{your}\mathrm{answere}\mathrm{is}\mathrm{true}$$

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