# 10th 7th chapter solution for trigonometry part3

This page 10th 7th chapter solution for trigonometry  part3 is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th 7th chapter solution for trigonometry part3

(ii) tan θ/(1-cot θ) + cot θ/(1-tanθ) = 1 + secθ cosecθ

Solution:

L.H.S

= tan θ/(1-cot θ) + cot θ/(1-tanθ)

a³ - b³ = (a - b) (a² + ab + b²)

= 1 + cosec θ sec θ

R.H.S

(iii) sin (90-θ)/(1-tanθ) + cos (90-θ)/(1-cotθ) = cosθ + sin θ

L.H.S

= sin (90-θ)/(1-tanθ) + cos (90-θ)/(1-cotθ)

= cos θ/(1-tanθ) + sin θ/(1-cotθ)

a² - b² = (a + b) (a - b)

= (cos θ + sin θ)(cos θ - sin θ)/(cos θ - sin θ)

= (cos θ + sin θ)

R.H.S

Hence proved

In the page 10th 7th chapter solution for trigonometry part3 we are going to see the solution of next problem

(iv) [tan (90-θ)/(cosecθ+1)]+[(cosecθ+1)/cotθ)] = 2 secθ

L.H.S

= [tan (90-θ)/(cosecθ+1)]+[(cosecθ+1)/cotθ)]

we can write tan (90-θ) as cot θ

=     (2/sinθ)/[Cos θ/sin θ]

=   (2/sin θ) x (sin θ/cos θ)

=    2/cos θ

=     2 sec θ

R.H.S

Hence proved

(v) (cotθ+cosecθ-1)/(cotθ-cosecθ+1)= cosecθ + cotθ

L.H.S

= (cotθ+cosecθ-1)/(cotθ-cosecθ+1)

R.H.S

Hence proved

(vi) (1 + cotθ - cosecθ) (1 + tanθ + secθ) = 2

L.H.S

=  2 sinθ cosθ/sinθ cosθ

= 2

R.H.S

Hence proved

(vii) (sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ-tanθ)

L.H.S

= (sinθ - cosθ + 1)/(sinθ + cosθ - 1)

dividing the whole equation by cos θ