Question

HEY GUYE!! GOOD EVENING!!
Prove that (sin A + cosec A)' +(cos A + sec A)' = 7+tan' A+ cot' A.​​

Answer 1

Step-by-step explanation:

Hey guys here your answer

Answer 2

Correct question:-

• $\sf{( \sin{A} + \cosec{A})^2 + ( \cos{A} + \sec{A})^2 = 7 + \tan^2{A} + \cot^2{A}}$

Solution:-

Identity used:-

$\bold{\underline{\boxed{\sf{\purple{\dag \; 1 + \tan^2{A} = \sec^2{A}}}}}}$

$\bold{\underline{\boxed{\sf{\purple{\dag \; 1 + \cot^2{A} = \cosec^2{A}}}}}}$

$\bold{\underline{\boxed{\sf{\purple{\dag \; \sin^2{A} + \cos^2{A} = 1}}}}}$

We know that,

★ $\sf{ \cosec{A} = \dfrac{1}{ \sin{A}}}$

and

★ $\sf{ \sec{A} = \dfrac{1}{ \cos{A}}}$

$\sf{( \sin{A} + \cosec{A})^2 + ( \cos{A} + \sec{A})^2 = 7 + \tan^2{A} + \cot^2{A}}$

Taking R.H.S :-

$\implies \sf{( \sin{A} + \cosec{A})^2 + ( \cos{A} + \sec{A})^2}$

$\implies \sf{ \sin^2{A} + \cosec^2{A} + 2 \sin{A} \cosec{A} + \cos^2{A} + \sec^2{A} + 2 \cos{A} \sec{A}}$

$\implies \sf{ \sin^2{A} + \cosec^2{A} + 2 \cancel{\sin{A}} \times \dfrac{1}{ \cancel{ \sin{A}}} + \cos^2{A} + \sec^2{A} + 2 \cos{A} \times \dfrac{1}{ \cancel{ \cos{A}}}}$

$\implies \sf{ \sin^2{A} + \cos^2{A} + \cosec^2{A} + \sec^2{A} + 2 + 2}$

$\implies \sf{1 + \cosec^2{A} + \sec^2{A} + 2 + 2}$

$\implies \sf{5 + (1 + \cot^2{A}) + (1 + \tan^2{A})}$

$\implies \sf{7 + \tan^2{A} + \cot^2{A}}$

$\rule{200}{2}$