Show that : (i) tan 48 tan 23 tan 42 tan 67 = 1 (ii) cos 38 cos 52 – sin 38 sin 52 = 0

Solution : (i)  L.H.S = tan 48 tan 23 tan 42 tan 67 = \(1\over cot 48\). tan 23 tan 42 \(1\over cot 67\) = \(1\over cot (90 – 42)\). tan 23 tan 42 \(1\over cot (90 – 23)\) = \(1\over tan 42\). tan 23 tan 42 \(1\over tan 23\) = 1 = R.H.S. (ii)  …

Show that : (i) tan 48 tan 23 tan 42 tan 67 = 1 (ii) cos 38 cos 52 – sin 38 sin 52 = 0 Read More »

State whether the following are true or false. Justify you answer.

Question : (i)  Sin (A + B) = sin A + sin B (ii)  The value of \(sin \theta\) increases as \(\theta\) increases. (iii)  The value of \(cos \theta\) increases as \(\theta\) increases. (iv)  \(sin \theta\) = \(cos \theta\) for all values of \(\theta\). (v)  Cot A is not defined for A = 0. Solution …

State whether the following are true or false. Justify you answer. Read More »

If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(1\over \sqrt{3}\) ; 0 < A + B \(\le\) 90 ; A \(\ge\) B, find A and B.

Solution : We have, tan (A + B) = \(\sqrt{3}\) \(\implies\)  tan (A + B) = tan 60 \(\implies\)  A + B = 60         …………(1) Also, tan (A – B) = \(1\over \sqrt{3}\) \(\implies\)  tan(A – B) = tan 30 \(\implies\)  A – B = 30            …………(2) …

If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(1\over \sqrt{3}\) ; 0 < A + B \(\le\) 90 ; A \(\ge\) B, find A and B. Read More »

State whether the following are true or false. Justify the answer.

Question : State whether the following are true or false. Justify the answer. (i)  The value of tan A is always less than 1. (ii)  sec A = \(12\over 5\) for some values of angle A. (iii)  cos A is the abbreviation used for the cosecant of angle A. (iv)  cot A is the product …

State whether the following are true or false. Justify the answer. Read More »

In \(\triangle\) PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution : We have, PQ = 5 cm PR + QR = 25 cm                  ………..(1) In triangle PQR, By Pythagoras Theorem, \({PR}^2\) = \({PQ}^2\) + \({QR}^2\) \(\implies\)  \({PQ}^2\) = \({PR}^2\) – \({QR}^2\) \(\implies\)  \({PQ}^2\) = (PR + QR)(PR – QR) \(\implies\)  \(5^2\) = (PR – QR). 25 …

In \(\triangle\) PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. Read More »