Question

please answer it.....​

Answer 1

Q4. (i)  We have,

$\displaystyle\longrightarrow\sf{\cot\theta=\dfrac{24}{7}}$

$\displaystyle\longrightarrow\sf{\tan\theta=\dfrac{7}{24}}$

By using the identity $\sf{\sec^2\theta=1+\tan^2\theta,}$

$\displaystyle\longrightarrow\sf{\sec^2\theta=1+\dfrac{49}{576}}$

$\displaystyle\longrightarrow\sf{\sec^2\theta=\dfrac{625}{576}}$

$\displaystyle\longrightarrow\sf{\sec\theta=\dfrac{25}{24}}$

$\displaystyle\longrightarrow\sf{\cos\theta=\dfrac{24}{25}}$

Now,

$\displaystyle\longrightarrow\sf{\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\sqrt{\dfrac{1-\dfrac{24}{25}}{1+\dfrac{24}{25}}}}$

$\displaystyle\longrightarrow\sf{\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\sqrt{\dfrac{25-24}{25+24}}}$

$\displaystyle\longrightarrow\sf{\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\sqrt{\dfrac{1}{49}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\dfrac{1}{7}}}}$

Q4. (ii)  We have,

$\displaystyle\longrightarrow\sf{4\cot\theta=5}$

$\displaystyle\longrightarrow\sf{4\cot\theta-5=0}$

$\displaystyle\longrightarrow\sf{4\cdot\dfrac{\cos\theta}{\sin\theta}-5=0}$

$\displaystyle\longrightarrow\sf{4\cos\theta-5\sin\theta=0}$

$\displaystyle\longrightarrow\sf{5\sin\theta-4\cos\theta=0}$

Then,

$\displaystyle\longrightarrow\sf{\dfrac{5\sin\theta+3\cos\theta}{5\sin\theta-2\cos\theta}=\dfrac{5\sin\theta-4\cos\theta+7\cos\theta}{5\sin\theta-4\cos\theta+2\cos\theta}}$

$\displaystyle\longrightarrow\sf{\dfrac{5\sin\theta+3\cos\theta}{5\sin\theta-2\cos\theta}=\dfrac{0+7\cos\theta}{0+2\cos\theta}}$

$\displaystyle\longrightarrow\sf{\dfrac{5\sin\theta+3\cos\theta}{5\sin\theta-2\cos\theta}=\dfrac{7\cos\theta}{2\cos\theta}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\dfrac{5\sin\theta+3\cos\theta}{5\sin\theta-2\cos\theta}=\dfrac{7}{2}}}}$

Q6. (i)  We see that,

$\longrightarrow\sf{\cos(2\theta)=\cos(\theta+\theta)}$

Since $\sf{\cos(A+B)=\cos A\cos B-\sin A\sin B,}$

$\longrightarrow\sf{\cos(2\theta)=\cos^2\theta-\sin^2\theta\quad\quad\dots(1)}$

$\longrightarrow\sf{\cos(2\theta)=\cos^2\theta+\sin^2\theta-2\sin^2\theta}$

$\longrightarrow\sf{\cos(2\theta)=1-2\sin^2\theta\quad\quad\dots(2)}$

On dividing (1) by (2),

$\longrightarrow\sf{\dfrac{\cos^2\theta-\sin^2\theta}{1-2\sin^2\theta}=\dfrac{\cos(2\theta)}{\cos(2\theta)}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{\cos^2\theta-\sin^2\theta}{1-2\sin^2\theta}=1}}}$

Q6. (ii)  From (1),

$\longrightarrow\sf{\cos(2\theta)=2\cos^2\theta-\cos^2\theta-\sin^2\theta}$

$\longrightarrow\sf{\cos(2\theta)=2\cos^2\theta-(\cos^2\theta+\sin^2\theta)}$

$\longrightarrow\sf{\cos(2\theta)=2\cos^2\theta-1\quad\quad\dots(3)}$

On dividing (3) by (1),

$\longrightarrow\sf{\dfrac{2\cos^2\theta-1}{\cos^2\theta-\sin^2\theta}=\dfrac{\cos(2\theta)}{\cos(2\theta)}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{2\cos^2\theta-1}{\cos^2\theta-\sin^2\theta}=1}}}$

Q7. Given,

$\longrightarrow\sf{\tan\theta+\dfrac{1}{\tan\theta}=2}$

Square both sides.

$\longrightarrow\sf{\left(\tan\theta+\dfrac{1}{\tan\theta}\right)^2=2^2}$

$\longrightarrow\sf{\tan^2\theta+\dfrac{1}{\tan^2\theta}+2\cdot\tan\theta\cdot\dfrac{1}{\tan\theta}=4}$

$\longrightarrow\sf{\tan^2\theta+\dfrac{1}{\tan^2\theta}+2=4}$

$\longrightarrow\sf{\underline{\underline{\tan^2\theta+\dfrac{1}{\tan^2\theta}=2}}}$