10th 7th chapter solution for trigonometry part2

This page 10th 7th chapter solution for trigonometry  part2 is going to provide you solution for every problems that you find in the exercise no 7.1

10th 7th chapter solution for trigonometry part2

(iv) cos θ/(sec θ - tan θ) = 1 + sin θ

Solution:

L.H.S

   = cos θ/(sec θ - tan θ)


(v) √( sec²θ + cosec²θ) = tan θ + cot θ

L.H.S

           = √( sec²θ + cosec²θ)

           = √(1 + tan²θ + 1 + cot²θ)

           = √(2 + tan²θ + cot²θ)

           = (tanθ)² + (cotθ)² + 2 tanθ x cotθ

           = (tanθ + cotθ)²

           = tanθ + cotθ


(vi) (1 + cos θ - sin²θ)/(sin θ)(1+cosθ) = cot θ

L.H.S

       = (1 + cos θ - sin²θ)/(sin θ)(1+cosθ)

       = (1 + cos θ) - (1 - cos²θ)/(sin θ)(1+cosθ)

       = [(1 + cos θ) - (1 - cos θ)(1 + cos θ)]/(sin θ)(1+cosθ)

       = [(1 + cos θ) (1 - (1 - cos θ))]/(sin θ)(1+cosθ)

taking
(1 + cos θ) as common term

       = [(1 + cos θ) (1 - 1 + cos θ))]/(sin θ)(1+cosθ)

       = [(1 + cos θ) (cos θ)]/(sin θ)(1+cosθ)

       = cos
θ/sin θ

       = cot θ

In the page 10th 7th chapter solution for trigonometry part2 we are going to see the solution of next problem

(vii) sec θ (1- sin θ)(sec θ + tan θ) = 1

L.H.S


(viii) sin θ/(cosec θ + cot θ) = 1 - cos θ

L.H.S

              = sin θ/(cosec θ + cot θ)

cot θ = cos θ/sin θ

              = sin θ/[(1/sin θ) + (cos θ/sin θ)]

              = sin θ/[(1+cosθ)/sin θ]

              = (sin θ x sin θ)/(1+cosθ)

              = sin²θ/(1+cosθ)

              = 1- cos²θ/(1+cosθ)

              = (1- cosθ)(1+cosθ)/(1+cosθ)

            = 1 - cos θ

               R.H.S


(3) Prove the following identities

(i) [sin (90-θ)/(1+sinθ)] + [cos θ/(1-(cos(90-θ))] = 2 sec θ

L.H.S

       = [sin (90-θ)/(1+sinθ)] + [cos θ/(1-(cos(90-θ))]

we can write sin (90-θ) as cos θ and cos (90 - θ) as sin θ

                = 2/cos θ

             = 2 sec θ




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