Question

Please answers it fast

Answer 1

Answer:

Which one u exactly wants??

Answer 2

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Solution 1 :-

➔(1- cos²A) cosec²A=1

➔as we know (1-cos²A)=sin²A

➔sin²A cosec²A

➔cosecA= $\frac{1}{sinA}$

➔so, sin²A × $\frac{1}{sin²A}$ = 1

Hence ,proved RHS =LHS

solution 2 :-

➔(1+cot²A)sin²A=1

➔as,we know (1+cot²A)=cosec²A

➔ cosec²A ×sin²A =1

Hence,RHS=LHS

solution 3 :-

➔tan²A cos²A=(1-cos²A)

➔as we know tan A =$\frac{sin A}{Cos A}$

➔$\frac{sin ²A}{Cos ²A}$ ×cos²A

➔ sin²A

➔as we know sin²A=(1-cos²A)

➔1-cos²A

➔Hence ,RHS=LHS

Solution 4:-

➔cosec√1-cos²A=1

➔cosec√sin²A

➔ cosecA ×SinA

➔$\frac{1}{sin A}$ ×sinA

➔1

Hence RHS=LHS

solution 5:-

➔(sec²-1)(cosec²-1)=1

➔( $\frac{1}{Cos ²A}$ -1) ( $\frac{1}{sin² A}$

➔$\frac{1-cos²A}{Cos 2A}$ ×$\frac{1-sin ²A}{sin² A}$

➔$\frac{sin² A}{Cos² A}$ ×$\frac{cos² A}{sin² A}$

➔1

➔Hence, RHS =LHS

solution 11 and 12 in attachment