Solution : The formula of tan (A + B) is \(tan A + tan B\over 1 – tan A tan B\). Proof : We have, tan (A + B) = \(sin (A + B)\over cos(A + B)\) Using sin (A+ B) and cos (A + B) formula, tan (A + B) = \(sin A cos B …
Solution : The value of cos 15 degrees is \(\sqrt{3} + 1\over 2\sqrt{2}\). Proof : We will write cos 15 as cos (45 – 30). By using formula cos (A – B) = cos A cos B + sin A sin B, cos (45 – 30) = cos 45 cos 30 + sin 45 sin 30 …
Solution : The value of sin 15 degrees is \(\sqrt{3} – 1\over 2\sqrt{2}\). Proof : We will write sin 15 as sin (45 – 30). By using formula sin (A – B) = sin A cos B – cos A sin B, sin (45 – 30) = sin 45 cos 30 – cos 45 sin …
Solution : The value of cos 75 degrees is \(\sqrt{3} – 1\over 2\sqrt{2}\). Proof : We will write cos 75 as cos (45 + 30). By using formula cos (A + B) = cos A cos B – sin A sin B, cos (45 + 30) = cos 45 cos 30 – sin 45 sin …
Solution : The value of sin 75 degrees is \(\sqrt{3} + 1\over 2\sqrt{2}\). Proof : We will write sin 75 as sin (45 + 30). By using formula sin (A + B) = sin A cos B + cos A sin B, sin (45 + 30) = sin 45 cos 30 + cos 45 sin …
Solution : In right angled triangle ABC, \(cosec \theta\) = \(AC\over BC\) \(\implies\) \(cosec^2 \theta\) = \(AC^2\over BC^2\) \(cot \theta\) = \(AB\over BC\) \(\implies\) \(cot^2 \theta\) = \(AB^2\over BC^2\) \(\implies\) 1 + \(cot^2 \theta\) = 1 + \(AB^2\over BC^2\) = \(BC^2 + AB^2\over BC^2\) = \(AC^2\over BC^2\) [ By Pythagoras theorem, \(AC^2\) = \(BC^2 + …
Prove that 1 + \(cot^2 \theta\) = \(cosec^2 \theta\). Read More »
Solution : In right angled triangle ABC, \(sec \theta\) = \(AC\over AB\) \(\implies\) \(sec^2 \theta\) = \(AC^2\over AB^2\) \(tan \theta\) = \(BC\over AB\) \(\implies\) \(tan^2 \theta\) = \(BC^2\over AB^2\) \(\implies\) 1 + \(tan^2 \theta\) = 1 + \(BC^2\over AB^2\) = \(AB^2 + BC^2\over AB^2\) = \(AC^2\over AB^2\) [ By Pythagoras theorem, \(AC^2\) = \(BC^2 + …
Prove that 1 + \(tan^2 \theta\) = \(sec^2 \theta\). Read More »
Solution : In right angled triangle ABC, \(sin \theta\) = \(BC\over AC\) \(\implies\) \(sin^2 \theta\) = \(BC^2\over AC^2\) \(cos \theta\) = \(AB\over AC\) \(\implies\) \(cos^2 \theta\) = \(AB^2\over AC^2\) On adding, \(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2\over AC^2\) + \(AB^2\over AC^2\) \(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2 + AB^2\over AC^2\) = \(AC^2\over AC^2\) …
Prove that \(sin^2 \theta\) + \(cos^2 \theta\) = 1. Read More »
Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\), then \(\angle\) C = 90 – \(\theta\) sec A = \(sec\theta\) = \(AC\over AB\) ……..(1) cosec C = \(cosec(90 – \theta)\) = \(AC\over AB\) …
Prove that \(Cosec(90 – \theta)\) = \(Sec\theta\). Read More »
Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\), then \(\angle\) C = 90 – \(\theta\) cosec A = \(cosec\theta\) = \(AC\over BC\) ……..(1) sec C = \(sec(90 – \theta)\) = \(AC\over BC\) …
Prove that \(Sec(90 – \theta)\) = \(Cosec\theta\). Read More »
