Question

10. Prove that (1+cote-cosece)(1+tan®+seco)=2​

Answer 1

Given:

• ( 1 + cot∅ - cosec∅ )(1+tan∅+sec∅) =2.

To Prove:

• LHS = RHS .

Proof:

LHS

= ( 1 + cot∅ - csc∅ )(1+tan∅+sec∅)

= ( 1 + cos∅/sin∅-csc∅)( 1 + sin∅/cos∅ + sec∅)

=( sin∅+cos∅-1/sin∅) ( cos∅+sin∅+1/cos∅)

= [(sin∅+cos∅-1)(sin∅+cos∅+1)] /sin∅.cos∅

= [{(sin∅+cos∅)-(1)}{(sin∅+cos∅)+(1)}] /sin∅.cos∅

= [ (sin∅+cos∅)²-(1)² ] / sin∅.cos∅

using

• (a+b)(a-b) = a²-b²

= [ sin²∅+cos²∅+2sin∅.cos∅-1]/ sin∅.cos∅

= [ 1 +2sin∅cos∅-1] / sin∅.cos∅

= 2sin∅.cos∅/sin∅.cos∅

= 2

= RHS

Hence proved .

Some more related formulae.

• sin²∅+cos∅ = 1
• sec²∅-tan²∅ = 1.
• cosec²∅-cot²∅ = 1.
• sin∅= 1/cosec∅
• cos∅=1/sec∅
• tan∅= 1/cot∅
• tan∅=sin∅/cos∅
• cot∅= cos∅/sin∅.

Answer 2

Step-by-step explanation:

Answer in attachment

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