Question 8 & 9, Review Exercise 10

Solutions of Question 8 & 9 of Review Exercise 10 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 the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.

We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{1}{2}\left[ 2\sin \left( \dfrac{\pi }{4}+\theta \right)\sin \left( \dfrac{\pi }{4}-\theta \right) \right]\\ &=\dfrac{1}{2}\left[ \cos \left( \left( \dfrac{\pi }{4}+\theta \right)-\left( \dfrac{\pi }{4}-\theta \right) \right)-\cos \left( \left( \dfrac{\pi }{4}+\theta \right)+\left( \dfrac{\pi }{4}+\theta \right) \right) \right]\\ &=\dfrac{1}{2}\left( \cos 2\theta -\cos \dfrac{\pi }{2} \right)\\ &=\dfrac{1}{2}\left( \cos 2\theta -0 \right)\\ &=\dfrac{\cos 2\theta }{2}=R.H.S.\end{align}

Prove that $\dfrac{{{\sin }^{2}}\left( \pi +\theta \right)\tan \left( \dfrac{3\pi }{2}+\theta \right)}{{{\cot }^{2}}\left( \dfrac{3\pi }{2}-\theta \right){{\cos }^{2}}\left( \pi -\theta \right)\cos ec\left( 2\pi -\theta \right)}=\cos \theta $.

\begin{align}L.H.S.&=\dfrac{{{\sin }^{2}}\left( \pi +\theta \right)\tan \left( \dfrac{3\pi }{2}+\theta \right)}{{{\cot }^{2}}\left( \dfrac{3\pi }{2}-\theta \right){{\cos }^{2}}\left( \pi -\theta \right)\cos ec\left( 2\pi -\theta \right)}\\ &=\dfrac{{{\left( \sin \left( 2\dfrac{\pi }{2}+\theta \right) \right)}^{2}}\tan \left( 3\dfrac{\pi }{2}+\theta \right)}{{{\left( \cot \left( 3\dfrac{\pi }{2}-\theta \right) \right)}^{2}}{{\left( \cos \left( 2\dfrac{\pi }{2}-\theta \right) \right)}^{2}}\cos ec\left( 2\pi -\theta \right)}\\ &=\dfrac{{{\left( -\sin \theta \right)}^{2}}\left( -co\operatorname{t}\theta \right)}{{{\left( tan\theta \right)}^{2}}{{\left( -\cos \theta \right)}^{2}}\cos ec\left( -\theta \right)}\\ &=\dfrac{{{\sin }^{2}}\theta \left( -\dfrac{\cos \theta }{\sin \theta } \right)}{\left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)\left( {{\cos }^{2}}\theta \right)\left( -\dfrac{1}{\sin \theta } \right)}\\ &=\cos \theta =R.H.S.\end{align}

Prove that $\dfrac{\cos \left( {{90}^{\circ }}+\theta \right)\sec \left( -\theta \right)\tan \left( {{180}^{\circ }}-\theta \right)}{\sec \left( {{360}^{\circ }}-\theta \right)\sin \left( {{180}^{\circ }}+\theta \right)\cot \left( {{90}^{\circ }}-\theta \right)}=-1$.

\begin{align}L.H.S.&=\dfrac{\cos \left( {{90}^{\circ }}+\theta \right)\sec \left( -\theta \right)\tan \left( {{180}^{\circ }}-\theta \right)}{\sec \left( {{360}^{\circ }}-\theta \right)\sin \left( {{180}^{\circ }}+\theta \right)\cot \left( {{90}^{\circ }}-\theta \right)}\\ &=\dfrac{\cos \left( {{90}^{\circ }}+\theta \right)\sec \theta \tan \left( 2\cdot {{90}^{\circ }}-\theta \right)}{\sec \left( 4\cdot {{90}^{\circ }}-\theta \right)\sin \left( 2\cdot {{90}^{\circ }}+\theta \right)\cot \left( {{90}^{\circ }}-\theta \right)}\\ &=\dfrac{\sin\theta \cdot \sec \theta \cdot \left( -\tan \theta \right)}{\sec \theta \left( \sin \theta \right)\tan \theta }\\ &=-1=R.H.S.\end{align}