Question

(sinθ–secθ)2+(cosθ–cosecθ)² = (1–secθ . cosecθ)²,Prove it by using trigonometric identities.

Answer 1

HELLO DEAR,

L.H.S.

= (sinθ + secθ)² + (cosθ + cosecθ)²

= sin²θ + 2sinθsecθ + sec²θ + cos²θ + 2cosθcosecθ + cosec²θ

=sin²θ + cos²θ + 2(sinθsecθ + cosθcosecθ) + sec²θ + cosec²θ

= 1 + 2(sinθ/cosθ + cosθ/sinθ) + (1/cos²θ + 1/sin²θ)

= 1 + 2(sin²θ + cos²θ)/sinθcosθ + (sin²θ + cos²θ)/sin²θcos²θ

= 1 + 2/sinθcosθ + 1/sin²θcos²θ

= 1 + 2secθcosecθ + sec²θcosec²θ

Now, R.H.S.

= ( 1 + secθcosecθ)²

= 1² + 2×1×(secθcosecθ) + (secθcosecθ)²

= 1 + 2secθcosecθ + sec²θcosec²θ

∴ L.H.S.= R.H.S.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi ,

Here I am using 'A' instead of theta

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1 ) sinA = 1/cosecA

2 ) cosA = 1/secA

3 ) sin²A + cos² A = 1

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LHS = ( sinA - SecA )² + ( cosA - cosecA )²

= ( 1/cosecA - secA )² + ( 1/secA - cosecA )²

= [(1-secAcosecA)/cosecA]²+[(1-secAcosecA)/secA]²

=( 1-secAcosecA)²[1/cosec²A + 1/sec²A ]

= ( 1 - secAcosecA )² [ sin²A + cos²A ]

= ( 1 - secAcosecA )²

= RHS

I hope this helps you.

: )