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




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