# 10th grade samacheer kalvi trigonometry

This page 10th grade samacheer kalvi trigonometry is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th grade samacheer kalvi trigonometry

(1) Determine whether each of the following is an identity or not.

(i) cos²θ + sec²θ = 2 + sin θ       Solution

(ii) cot²θ + cos θ = sin² θ          Solution

(2) Prove the following identities

(i) sec²θ + cosec²θ = sec²θ  cosec²θ  Solution

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

(iii) (1-sin θ)/(1+sin θ) = sec θ - tan θ   Solution

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

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

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

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

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

(3) Prove the following identities

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

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

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

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

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

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

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

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

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

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

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