Question

prove that (sin A+cosec A)^2 + (cos A+secA)^2 =7+ tan^2 A + cot^2A. ​

Answer 1

Answer:

Step-by-step explanation:

⇒ ( sinA + cosecA )² + ( cosA + secA )²

⇒ ( sin²A + cosec²A + 2sinAcosecA ) + ( cos²A + sec²A + 2cosAsecA )

        sinAcosecA = 1 = cosAsecA

⇒ ( sin²A + cosec²A + 2(1) ) + ( cos²A + sec²A + 2(1) )

⇒ sin²A + cosec²A + 2 + cos²A + sec²A + 2

⇒ sin²A + cos²A + 2 + 2 + cosec²A + sec²A

       cosec²A = 1 + cot²A ; sec²A = 1 + tan²A

⇒ 1 + 2 + 2 + ( 1 + cot²A ) + ( 1 + tan²A )

⇒ 1 + 2 + 2 + 1 + cot²A + 1 + tan²A

⇒  7 + tan²A + cot²A

  Hence proved.

Answer 2

To Prove :-

→ (sin A + Cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A .

Solution :-

Taking LHS -

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

$\sf{ \implies \: \cosec A = \frac{1}{ \sin A} \: and \: \sec A = \frac{1}{ \cos A} }$

$\sf{ \implies \: ( { \sin A + \frac{1}{ \sin A} )}^{2} + ( { \cos A + \frac{1}{ \cos A } }^{2} }$

$\sf{\implies \: ( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

Using this identity

$\sf{ \implies \: { \sin }^{2} A + \frac{1}{ { \sin }^{2} A} + 2 \sin A. \frac{1}{ \sin A } + { \cos }^{2} A + \frac{1}{ { \cos }^{2} A} + 2 \cos A. \frac{1}{ \cos A } }$

$\sf{\implies \: {\sin }^{2} A + \frac{1}{{\sin }^{2} A} +2 + { \cos }^{2} A + \frac{1}{ { \cos }^{2} A} + 2 }$

sin²A + cos²A = 1

$\sf{\implies \: 2 + 2 + 1 + \frac{1}{ { \sin }^{2} A} + \frac{1}{ { \cos }^{2}A } }$

Putting value of 1 in fraction

$\sf{ \implies \: 5 + \frac{ { \sin }^{2} A + { \cos }^{2} A}{ { \sin }^{2} A } + \frac{ { \sin }^{2} A + { \cos }^{2} A}{ { \cos }^{2} A} }$

$\sf{ \implies \: 5 + \frac{ { \sin }^{2}A }{ { \sin }^{2} A} + \frac{ { \cos }^{2}A }{ { \sin }^{2} A } + \frac{ { \sin }^{2} A}{ { \cos }^{2} A } + \frac{ { \cos }^{2} A }{ { \cos }^{2} A }}$

$\sf{\implies \: 5 + 1 + 1 + \frac{ { \sin }^{2} A}{ { \cos }^{2} A} + \frac{ { \cos }^{2} A }{ { \sin }^{2} A }}$

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

LHS = RHS

Hence proved