Question

I will mark correct explained answer as branliest answer
​

Answer 1

Correct Question :--

Prove that :-- (sin@-cos@/sin@+cos@) + (sin@+cos@/sin@-cos@) = 2/(2sin²@ - 1)

[ in RHS , denominator we also have 2 multiply by sin²@ than only our Question will be correct and we can prove this . ] .

See solution now :---

Taking LHS we get,

→ (sin@-cos@/sin@+cos@) + (sin@+cos@/sin@-cos@)

Taking lcm of denominator we get,

→ [ (sin@-cos@)² + (sin@+cos@)²] / [(sin@+cos@)(sin@-cos@)]

Now, using (a-b)² = a² + b² - 2ab and (a+b)² = a² + b² - 2ab in Numerator , and (a+b)(a-b) = a² - b² in denominator we get,

→ [ sin²@ + cos²@ - 2sin@*cos@] + [ sin²@ + cos²@ - 2sin@*cos@] / [ sin²@ - cos²@ ]

Adding Numerator part now we get,

→ 2sin²@ + 2cos²@ / [ sin²@ - cos²@ ]

→ 2[ sin²@ + cos²@ ] / [ sin²@ - cos²@ ]

Now, using sin²@+cos²@ = 1 in Numerator and cos²@ = (1-sin²@) in Denominator we get,

→ 2/sin²@ - (1-sin²@)

→ 2/(2sin²@ - 1) = RHS

✪✪ Hence Proved ✪✪

Answer 2

Answer:

Correct Question :--

Prove that :-- (sin@-cos@/sin@+cos@) + (sin@+cos@/sin@-cos@) = 2/(2sin²@ - 1)

[ in RHS , denominator we also have 2 multiply by sin²@ than only our Question will be correct and we can prove this . ] .

See solution now :---

Taking LHS we get,

→ (sin@-cos@/sin@+cos@) + (sin@+cos@/sin@-cos@)

Taking lcm of denominator we get,

→ [ (sin@-cos@)² + (sin@+cos@)²] / [(sin@+cos@)(sin@-cos@)]

Now, using (a-b)² = a² + b² - 2ab and (a+b)² = a² + b² - 2ab in Numerator , and (a+b)(a-b) = a² - b² in denominator we get,

→ [ sin²@ + cos²@ - 2sin@*cos@] + [ sin²@ + cos²@ - 2sin@*cos@] / [ sin²@ - cos²@ ]

Adding Numerator part now we get,

→ 2sin²@ + 2cos²@ / [ sin²@ - cos²@ ]

→ 2[ sin²@ + cos²@ ] / [ sin²@ - cos²@ ]

Now, using sin²@+cos²@ = 1 in Numerator and cos²@ = (1-sin²@) in Denominator we get,

→ 2/sin²@ - (1-sin²@)

→ 2/(2sin²@ - 1) = RHS