Question

SOLVE THIS QUESTIONS PLEASE.

WILL BE MARKED AS BRAINLIEST...

Answer 1

Hii !!

Solution is here chote !!

(1 ) from LHS

(1 - sin²¢ ) sec²¢

= cos²¢ × sec²¢

= cos²¢ × 1/cos²¢ = 1 RHS proved

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(2)

from LHS ,

( 1 + tan²¢ ) cos²¢

= sec²¢ × cos²¢

= 1/cos²¢ × cos²¢ = 1 RHS prooved

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(3)

From LHS

(1 + tan²¢ ) ( 1 + sin¢ ) ( 1 - sin¢ ) ( question misstyped)

= sec²¢ ( 1 - sin²¢ )

= sec²¢ × cos²¢

= sec²¢ × 1/sec²¢ = 1 RHS prooved

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(4)

from LHS

{(2cos²¢ + 2/( 1 + cot²¢ )}


= 2cos²¢ + 2/ cosec²¢

= 2cos²¢ + 2sin²¢

= 2 ( cos²¢ + sin²¢ )

= 2 RHS prooved

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Hope it helps you !!!

@muskraj❤

Answer 2

$using \: \: identity \: \: 1 - { \sin( \alpha ) }^{2} = ){ \cos( \alpha ) }^{2}$

$we \: get \:$
1)
${ \cos( \alpha ) }^{2} \times { \sec( \alpha ) }^{2} = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ as \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( \cos( \alpha ) \times \sec( \alpha ) ) \\ hence \: proof$

$now \: 2$

$using \: identity \: 1 + { \tan( \alpha ) }^{2} = { \sec( \alpha ) }^{2}$

$we \: get$

${ \sec( \alpha ) }^{2} \times { \cos( \alpha ) }^{2} = 1 \: \: \: hence \: proof$