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- Question 7, Exercise 10.2
- n} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p5|< Question 6]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p7|Question 8, 9 >]]</btn></text>
- Question 8 and 9, Exercise 10.2
- -1}=R.H.S.\end{align} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p6|< Question 7]]</btn></text>
- Question 6, Exercise 10.2
- <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p4|< Question 4, 5]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p6|Question 7 >]]</btn></text>
- Question 2, Exercise 10.1
- ====== Question 2, Exercise 10.1 ====== Solutions of Question 2 of Exercise 10.1 of Unit 10: Trigonometric Identities... (KPTB or KPTBB) Peshawar, Pakistan. ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\
- Question 13, Exercise 10.1
- } \text{ and } r=\sqrt{2}.\end{align} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-1-p10|< Question 11,12]]</btn></text>
- Question 5, Exercise 10.3
- stan. =====Question 5(i)===== Prove that $$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ ... \dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta ... ha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2
- Question 2, Exercise 10.3
- ====== Question 2, Exercise 10.3 ====== Solutions of Question 2 of Exercise 10.3 of Unit 10: Trigonometric Identities ... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 2(i)===== Convert the sum or difference as product:... We have an identity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos
- Question 5, Exercise 10.3
- istan. =====Question 5(i)===== Prove that $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ ... =\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta ... ha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2
- Question 3, Exercise 10.1
- n} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-1-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-1-p4|Question 4 >]]</btn></t
- Question 6, Exercise 10.1
- ====Question 6(i)===== Show that: $\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}$ ====Solution==== \begin{align}L.H.S&=\cos \alpha \\ \cos \al
- Question11 and 12, Exercise 10.1
- ====== Question11 and 12, Exercise 10.1 ====== Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric Identi... a triangle $ABC$, show that $\cot \dfrac{\alpha }{2}+\cot \dfrac{\beta }{2}+\cot \dfrac{\gamma }{2}=\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2}\cot \d
- Question 4 and 5, Exercise 10.2
- }$$ <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p3|< Question 3]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p5|Question 6 >]]</btn></text>
- Question 2, Exercise 10.2
- }$$ <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p1|< Question 1]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p3|Question 3 >]]</btn></text>
- Question 8 & 9, Review Exercise 10
- \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-... ft( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{1}{2}\left[ 2\sin \left( \dfrac{\pi }{4}+\theta \righ
- Question 5, Exercise 10.1
- lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alp... ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqr