Question

119. If seco + tano = 2 + √5 then the value
of sino + coso ?​

Answer 1

Question:

If sec∅ + tan∅ = √5 + 2 , then find the value of sin∅ + cos∅ .

Answer:

sin∅ + cos∅ = 3/√5 or 3√5/5

Note:

• sin²∅ + cos²∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

• sec∅ = 1/cos∅

• cosec∅ = 1/sin∅

• tan∅ = sin∅/cos∅

• cot∅ = 1/tan∅ = cos∅/sin∅

• (A+B)² = A² + B² + 2•A•B

• (A-B)² = A² + B² - 2•A•B

• (A+B)•(A-B) = A² - B²

Solution:

Given: sec∅ + tan∅ = √5 + 2

To find : sin∅ + cos∅ = ?

Let,

sec∅ + tan∅ = √5 + 2 ------(1)

Also,

We know that,

=> sec²∅ - tan²∅ = 1

=> (sec∅ - tan∅)•(sec∅ + tan∅) = 1

=> sec∅ - tan∅ = 1/(sec∅ + tan∅)

=> sec∅ - tan∅ = 1/(√5 + 2) { using eq-(1) }

Now,

Rationalising the denominator in LHS, we have ;

=> sec∅ - tan∅ = (√5 - 2)/[(√5 + 2)•(√5 - 2)]

=> sec∅ - tan∅ = (√5 - 2)/[(√5)² - 2²]

=> sec∅ - tan∅ = (√5 - 2)/(5 - 4)

=> sec∅ - tan∅ = √5 - 2 ---------(2)

Now,

Adding eq-(1) and eq-(2) , we have ;

=> sec∅ + tan∅ + sec∅ - tan∅ = √5 + 2 + √5 - 2

=> 2•sec∅ = 2√5

=> sec∅ = 2√5/2

=> sec∅ = √5

=> 1/cos∅ = √5

=> cos∅ = 1/√5

Also,

We know that ,

=> sin²∅ + cos²∅ = 1

=> sin²∅ = 1 - cos²∅

=> sin²∅ = 1 - (1/√5)²

=> sin²∅ = 1 - 1/5

=> sin²∅ = (5-1)/5

=> sin²∅ = 4/5

=> sin∅ = √(4/5)

=> sin∅ = 2/√5

Thus,

=> sin∅ + cos∅ = 2/√5 + 1/√5

=> sin∅ + cos∅ = (2+1)/√5

=> sin∅ + cos∅ = 3/√5 or 3√5/5

Hence,

The required value of sin∅ + cos∅ is 3/√5 or 3√5/5 .