# 10th 7th chapter solution for trigonometry part4

This page 10th 7th chapter solution for trigonometry  part4 is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th 7th chapter solution for trigonometry part4

(viii) tan θ/(1-tan²θ)= sinθsin(90-θ)/[2sin²(90-θ) - 1]

Solution:

L.H.S

= tan θ/(1-tan²θ)

= sin θ sin (90-θ)/[2sin²(90-θ) - 1]

R.H.S

Hence proved

(ix) [1/(cosecθ-cotθ)]-(1/sinθ) = [(1/sinθ)]-[1/(cosecθ+cotθ)]

L.H.S

= [1/(cosecθ-cotθ)]-(1/sinθ)

multiplying the first fraction by (cosecθ+cotθ)

= [(1/(cosecθ-cotθ)) x (cosecθ+cotθ)/(cosecθ+cotθ)]-      (1/sinθ)

= (cosecθ+cotθ)/(cosec²θ-cot²θ)-(1/sinθ)

= [(cosecθ+cotθ)/1]-(1/sinθ)

= [(cosecθ+cotθ) - cosec θ]

= cot θ  -----(1)

= (1/sinθ) - [1/(cosecθ-cotθ)]

multiplying the second fraction by (cosecθ+cotθ)

=  [(cosecθ-cotθ)/(cosecθ-cotθ)(cosecθ+cotθ)] - (1/sinθ)

= (cosecθ+cotθ)/(cosec²θ-cot²θ)-(1/sinθ)

= [(cosecθ+cotθ)/1]-(1/sinθ)

= [(cosecθ+cotθ) - cosec θ]

= cot θ  -----(2)

(1) = (2)

Hence proved

In the page 10th 7th chapter solution for trigonometry part4 we are going to see the solution of next problem

(x) (cot²θ+sec²θ)/(tan²θ+cosec²θ)=sinθcosθ(tanθ+cotθ)

L.H.S

= (cot²θ+sec²θ)/(tan²θ+cosec²θ)

= (cosec²θ - 1 + 1+ tan²θ)/(tan²θ+cosec²θ)

= (tan²θ+cosec²θ)/(tan²θ+cosec²θ)

= 1 -----(1)

R.H.S

= sinθcosθ (tanθ+cotθ)

= sinθcosθ [(sinθ/cosθ)+(cosθ/sinθ)]

taking L.C.M

= sinθcosθ [(sin²θ+cos²θ)/(cosθsinθ)]

=  sinθcosθ [1/(cosθsinθ)]

= 1-----(2)

(1) = (2)

Hence proved

(4) If x = a secθ + b tan θ and y = a tan θ+ b secθ then prove that x² - y² = a² - b²

Solution:

x = a secθ + b tan θ

x² =  (a secθ + b tan θ)²

=  (a secθ)² + (b tan θ)² + 2 a b sec θ tan θ

=  a² sec²θ + b² tan² θ + 2 a b sec θ tan θ  --(1)

y = a tan θ + b sec θ

y² =  (a tan θ + b sec θ)²

=  (a tanθ)² + (b sec θ)² + 2 a b sec θ tan θ

=  a² tan²θ + b² sec² θ + 2 a b sec θ tan θ --(2)

x² - y² = (a² sec²θ + b² tan² θ + 2 a b sec θ tan θ) -
(a² tan²θ + b² sec² θ + 2 a b sec θ tan θ)

= a² sec²θ - a² tan²θ + b² tan²θ - b² sec²θ +
2 a b sec θ tan θ  - 2 a b sec θ tan θ

= a²(sec²θ - tan²θ ) + b²(tan²θ -sec²θ)

= a²(secc²θ - tan²θ ) -  b²(sec²θ -  tan²θ )

= a²(1 ) -  b²(sec²θ -  tan²θ )