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Standard Values of Trigonometric Ratios

tan 2 A / 1+t...

Question

tan²A/1+tan²A + cot²A/1+cot²A=sec²Acos²A-2secAcosA

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Solution

When we take -ve outside on RHS it becomes
-(-sec^2Acos^A+)
=-(-1+2)
=-(1)
=-1

It seems your question is incorrect
Qn should be like( 2secAcosA-sec^2Acos^A) on RHS

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\frac{{tan}^{2} A}{1 + {tan}^{2} A} + \frac{{cot}^{2} A}{1 + {cot}^{2} A} = 1

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - {tan}^{2} A}{1 - {cot}^{2} A})}^{2} = {tan}^{2} A

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {tan}^{2} A

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - {tan}^{2} A}{1 - {cot}^{2} A})}^{2} = {tan}^{2} A

\frac{1 - {tan}^{2} A}{{cot}^{2} A - 1} = {tan}^{2} A

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