Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

The value of the expression:

$$bc+\frac{1}{ck}+\frac{ak}{1+bk}$$

is equal to:

$$\frac{1}{k}\left(a+\frac{1}{a}\right)$$

***

Step 1: Express trigonometric functions in terms of given variables and the constant $k$.

From the problem statement, we have the following relationships:

* $\sin x=ak$

* $\cos x=bk$

* $\tan x=ck$

Step 2: Establish a relationship between $b$, $c$, $a$, and $k$.

Using the identity $\tan x=\frac{\sin x}{\cos x}$, we can write:

$$\frac{c}{k}=\frac{a/k}{b/k}$$

$$c=\frac{a}{b}$$

which implies $bc=a$.

***

Step 3: Substitute the expressions into the given equation and simplify.

Substitute the expressions for $b$, $c$, $\sin x$, and $\cos x$ into the expression $bc+\frac{1}{ck}+\frac{ak}{1+bk}$:

* $bc=a$

* $\frac{1}{ck}=\frac{1}{\tan x}=\cot x=\frac{\cos x}{\sin x}$

* $\frac{ak}{1+bk}=\frac{\sin x}{1+\cos x}$

The full expression becomes:

$$bc+\frac{1}{ck}+\frac{ak}{1+bk}=a+\cot x+\frac{\sin x}{1+\cos x}$$

Now, use half-angle formulas to simplify:

* $\cot x=\frac{\cos x}{\sin x}=\frac{{\cos }^2(x/2)-{\sin }^2(x/2)}{2\sin (x/2)\cos (x/2)}=\frac{\cos (x/2)}{2\sin (x/2)}-\frac{\sin (x/2)}{2\cos (x/2)}=\frac{1}{2}\cot (x/2)-\frac{1}{2}\tan (x/2)$

* $\frac{\sin x}{1+\cos x}=\frac{2\sin (x/2)\cos (x/2)}{2{\cos }^2(x/2)}=\frac{\sin (x/2)}{\cos (x/2)}=\tan (x/2)$

Alternatively, combine the trigonometric terms:

$$a+\frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x}=a+\frac{\cos x(1+\cos x)+{\sin }^{2}x}{\sin x(1+\cos x)}$$

$$=a+\frac{\cos x+{\cos }^{2}x+{\sin }^{2}x}{\sin x(1+\cos x)}$$

$$=a+\frac{\cos x+1}{\sin x(1+\cos x)}=a+\frac{1}{\sin x}$$

Step 4: Substitute back in terms of $a$ and $k$.

Substitute $\sin x=ak$ into the simplified expression:

$$a+\frac{1}{ak}$$

$$a+\frac{1}{ak}=\frac{a^{2}k+1}{ak}=\frac{1}{k}\left(a+\frac{1}{a}\right)$$