Question

please help me

i don't known the problem
please answer me with step
this problem difficult
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Answer 1

Answer:

Step-by-step explanation:

Here I am using A instead of theta.

LHS = secA/(secA+1) + secA/(secA-1)

= SecA[ 1/(secA+1) + 1/(secA-1 ) ]

= SecA[ ( secA-1+secA+1 )/(sec² A-1²) ]

= SecA[ 2secA/tan² A ]

= 2sec²A/ tan² A

= ( 2/cos² A ) / ( sin² A/cos² A )

= 2/sin² A

= 2cosec²A

= RHS

I hope this helps you.

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

Prove :- secA/(secA+1) + secA/(secA - 1) = 2cosec²A

Sᴏʟᴜᴛɪᴏɴ :-

Solving LHS, by taking LCM we get,

→ secA/(secA+1) + secA/(secA - 1)

→ [secA(secA - 1) + secA(secA + 1)] / [(secA+1)(secA - 1)]

Now using (a + b)(a - b) in Denominator and solving Numerator, we get,

→ [sec²A - secA + sec²A + secA] / (sec²A - 1)

→ 2sec²A/(sec²A - 1)

Now using sec²A - 1 = tan²A in Denominator, we get,

→ 2sec²A/tan²A

Now using secA = 1/cosA & tanA = sinA/cosA we get,

→ {2(1/cos²A)} /{sin²A/cos²A}

→ (2/cos²A) * (cos²A/sin²A)

→ 2/sin²A

in last using (1/sinA) = cosecA , we get,

→ 2cosec²A = RHS (Proved).