Question

Prove the identity (sinA + cosA) (tanA + cotA) = secA + CosecA

Answer 1

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Answer 2

Step-by-step explanation:

$(\sin A+\cos A)(\tan A+\cot A)=\sec A+\csc A$

As we all know that $\tan A=\frac{(\sin A)}{(\cos A)} \text { and } \cot A=\frac{(\cos A)}{(\sin A)}$, therefore putting the value of tan and cot in term of cos and sine.

$(\sin A+\cos A)\left(\frac{(\sin A)}{(\cos A)}+\frac{(\cos A)}{(\sin A)}\right)=\sec A+\csc A$

$(\sin A+\cos A)\left(\frac{\sin ^{2} A+\cos ^{2} A}{\cos A \cdot \sin A}\right)=\sec A+\csc A$

After simplification we can see that $\sin ^{2} A+\cos ^{2} A$ is the numerator as we all know that $\sin ^{2} A+\cos ^{2} A=1$

$(\sin A+\cos A)\left(\frac{1}{\cos A \cdot \sin A}\right)=\sec A+\csc A$

After putting, $\sin ^{2} A+\cos ^{2} A=1$, we multiply $(\sin A+\cos A)$ to form the sin and cos in terms of sec and cosec.

$(\sin A+\cos A)(\tan A+\cot A)=\left(\frac{\sin A}{\cos A \sin A}+\frac{\cos A}{\cos A \sin A}\right)$

$=\left(\frac{1}{\cos A}+\frac{1}{\sin A}\right)$

$\therefore(\sin A+\cos A)(\tan A+\cot A)=\sec A+\csc A$

Hence proved