Trigonometry Questions

Solution : The formula of cot (A – B) is \(cot A cot B + 1\over cot B – cot A\). Proof : We have, cot (A – B) = \(cos (A – B)\over sin(A – B)\) Using sin (A – B) and cos (A – B) formula, cot (A – B) = \(cos A cos […]

Solution : The formula of cot (A + B) is \(cot A cot B – 1\over cot B + cot A\). Proof : We have, cot (A + B) = \(cos (A + B)\over sin(A + B)\) Using sin (A + B) and cos (A + B) formula, cot (A + B) = \(cos A cos

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

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 +

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 +