Question 9 and 10, Exercise 10.1

Solutions of Question 9 and 10 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that: $\dfrac{\sin \theta }{\sec 4\theta }+\dfrac{\cos \theta }{\cos ec4\theta }=\sin 5\theta$

$$\begin{align}L.H.S.&=\dfrac{\sin \theta }{\sec 4\theta }+\dfrac{\cos \theta }{\cos ec4\theta }\\ &=\dfrac{\sin \theta }{\frac{1}{\cos 4\theta }}+\dfrac{\cos \theta }{\frac{1}{\sin 4\theta }}\\ &=\sin \theta \cos 4\theta +\cos \theta \sin 4\theta \\ &=\sin \left( \theta +4\theta \right)\\ &=\sin 5\theta =R.H.S.\end{align}$$

Show that: $\dfrac{\sin \left( {{180}^{\circ }}-\alpha \right)\cos \left( {{270}^{\circ }}-\alpha \right)}{\sin \left( {{180}^{\circ }}+\alpha \right)\cos \left( {{270}^{\circ }}+\alpha \right)}=1$

$$\begin{align}L.H.S.&=\dfrac{\sin \left( {{180}^{\circ }}-\alpha \right)\cos \left( {{270}^{\circ }}-\alpha \right)}{\sin \left( {{180}^{\circ }}+\alpha \right)\cos \left( {{270}^{\circ }}+\alpha \right)}\\ &=\dfrac{\left( \sin {{180}^{\circ }}\cos \alpha -\cos {{180}^{\circ }}\sin \alpha \right)\left( \cos {{270}^{\circ }}\cos \alpha -\sin {{270}^{\circ }}\sin \alpha \right)}{\left( \sin {{180}^{\circ }}\cos \alpha +\cos {{180}^{\circ }}\sin \alpha \right)\left( \cos {{270}^{\circ }}\cos \alpha +\sin {{270}^{\circ }}\sin \alpha \right)}\\ &=\dfrac{\left( \left( 0 \right)\cos \alpha -\left( -1 \right)\sin \alpha \right)\left( \left( 0 \right)\cos \alpha -\left( -1 \right)\sin \alpha \right)}{\left( \left( 0 \right)\cos \alpha +\left( -1 \right)\sin \alpha \right)\left( \left( 0 \right)\cos \alpha +\left( -1 \right)\sin \alpha \right)}\\ &=\dfrac{\left( \sin \alpha \right)\left( \sin \alpha \right)}{\left( -\sin \alpha \right)\left( -\sin \alpha \right)}\\ &=1=R.H.S.\end{align}$$

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