Question

(sin A + sec A)^2 + (cos A + cosec A)^2 = (1 + sec A cosec A)^2​

Answer 1

Answer:

i think this will help you...

Step-by-step explanation:

( sin a +sec a)2+( cos a+ cosec a)2= (1+sec a cosec a)2

(Sin a + 1/cos a)2 + (cos a +1/sin a)2

=) (sin a cos a+1)2+ (cos a sin a+1)2

Cos2asin2a

=) (sin a cos a+1)2( 1/cos2a + 1/sin2a)

=) (sin a cos a+1)2(sin2a + cos2a)

Sin2a cos2a

=)(sin a cos a+1)2

Sin2a cos2a

=)sin2a cos2a +1+2sin a cos a=)sin2a cos2a+1+2sin a cos a

Sin2a cos2aSin2a cos2aSin2a cos2aSin2a cos2a

=) 1+11+2

Sin2acos2asin a cos a

=) 1+ cosec2a sec2a + 2cosec a sec a

=) (1+ cosec a sec a)2

Hence proved.

Or...

this is also a easy one...

( sin a + sec a)2+ ( cos a + cosec a)2

= ( 1 / cosec a + sec a)2+ ( 1 / sec a + cosec a)2

= (1 + sec a cosec a / cosec a)2+ ( 1 + sec a cosec a / sec a )2

=( 1 + sec a cosec a)2{ 1 / cosec2a + 1 / sec2a ) [ taking ( 1 + sec a cosec a)2as common]

= ( 1 + sec a cosec a)2( sin2a + cos2a )

= ( 1 + sec a cosec a)2( 1) (since sin2a + cos2a = 1 )

= ( 1 + sec a cosec a)2hence prooved.

Answer 2

$( {sinA + secA})^{2} + (cosA + cosecA) ^{2} \\ \\ = ( {sinA + \frac{1}{cosA} })^{2} + ( {cosA + \frac{1}{sinA} })^{2} \\ \\ = {sin}^{2} A + \frac{1}{ {cos}^{2} A} + 2 \frac{sinA}{cosA} + {cos}^{2} A + \frac{1}{ {sin}^{2} A} + 2 \frac{cosA}{sinA} \\ \\ = 1 + ( \frac{1}{ {sin}^{2}A } + \frac{1}{ {cos}^{2} A} ) + 2( \frac{sinA}{cosA} + \frac{cosA}{sinA} ) \\ \\ = 1 + \frac{1}{ {sin}^{2} A {cos}^{2}A } + 2 (\frac{1}{sinAcosA} ) \\ \\ = 1 + {sec}^{2} A {cosec}^{2} A + 2secAcosecA \\ \\ = ( {1 + secAcosecA})^{2}$