Question

please experts answer this but dont spam; answer is 7

Answer 1

the answer that i came up with is 7, do you need me to show work? hope you don't mind that I used X instead of a.

(sin X + cosec X)^2+(cos X + sec X)^2

sin^2 X + cosec^2 X + 2*sin X*cosec X + cos^2 + sec^2 X + 2 * cos C * sec X

1 + cosec^2 X + 2 + sec^2 X + 2

1 + 1 + cot^2 X + 2 + 1 + tan2^ X + 2

7 + cot^2 X + tan^2 X

this proves that

(sin X + cosec X)^2 + (cos X + sec X)^2 = 7 + cot^2 X + tan^2 X

Answer 2

Given,
$(sin\theta + cosec\theta)^2 + (cos\theta +sec\theta)^2 = k + tan^2 \theta + cot^2 \theta$



Note : theta is written as A.


Solving left hand side of the given equation.


= > ( sinA + cosecA )² + ( cosA + secA )²



==============
We know,
( a + b )² = a² + b² + 2ab
==============



= > ( sin²A + cosec²A + 2 sinA cosecA ) + ( cos²A + sec²A + 2 cosA secA )



=============
sinA = 1 / cosecA
cosA = 1 / secA
=============



= > sin²A + cosec²A + 2( sinA x 1 / sinA ) + cos²A + sec²A + 2( secA x 1 / secA )


= > sin²A + cosec²A + 2( 1 ) + cos²A + sec²A + 2( 1 )


= > sin²A + cosec²A + 2 + cos²A + sec²A + 2


= > ( sin²A + cos²A ) + ( cosec²A ) + ( sec²A ) + 2 + 2



============
sin²A + cos²A = 1
cosec²A = 1 + cot²A
sec²A = 1 + tan²A
============



= > ( 1 ) + ( 1 + cot²A ) + ( 1 + tan²A ) + 4


= > 1 + 1 + cot²A + 1 + tan²A + 4


= > 7 + cot²A + tan²A



Now, comparing left and right hand sides,

= > 7 + cot²A + tan²A = k + cot²A + tan²A


= > 7 + cot²A - cot²A + tan²A - tan²A = k


= > 7 = k



Therefore the numeric value of used variable'x' is 7
$\:$