Find the value of (1+tan 245") (1 + tan 250) (1 + tan 260) (1-tan 200") (1-tan 205") (1-tan 215") 28. (A)6 (B) 8 (C) 9 (D) 10

Key ideas you must know

1. Periodicity of tangent
   $\tan(\theta+180^\circ)=\tan\theta$ This lets us bring all angles inside $0^\circ$ to $90^\circ$ so that the numbers are friendlier.

2. Complementary-angle (co-function) identity
   $\tan\bigl(90^\circ-\theta\bigr)=\cot\theta=\frac{1}{\tan\theta}$

3. A smart algebraic product
   For any $t=\tan\theta$: [ \bigl(1+\cot\theta\bigr)\bigl(1-\tan\theta\bigr)=\bigl(1+\frac{1}{t}\bigr)(1-t)=\frac{(1+t)(1-t)}{t}=\frac{1-t^2}{t}. ] Notice how two separate brackets merge into one simpler fraction.

Logical chain to tackle the problem

1. Convert each angle to an acute one (<90°).
   For instance, $245^\circ\to65^\circ$, $250^\circ\to70^\circ$, $260^\circ\to80^\circ$, etc.
2. Group complementary partners.
   $65^\circ$ pairs with $25^\circ$ (because $65+25=90$), and $70^\circ$ pairs with $20^\circ$.
3. Apply the identity above to each pair so each duo collapses to a single, shorter factor.
4. Multiply the three simpler factors you are left with—this is just ordinary number-crunching. A small pocket-approximation or a bit of known tangent tables gives the answer quickly.

This combination of periodicity + co-functions + pair-wise simplification is the standard JEE trick whenever you see long chains of $1\pm\tan(\text{angle})$ terms.

What’s the question?

We have to multiply six brackets full of tan numbers:

• $(1+\tan245^\circ)$
• $(1+\tan250^\circ)$
• $(1+\tan260^\circ)$
• $(1-\tan200^\circ)$
• $(1-\tan205^\circ)$
• $(1-\tan215^\circ)$

How should we look at it?

1. Tangent repeats every $180^\circ$. So $\tan(245^\circ)=\tan(65^\circ)$ because $245-180=65$.
2. $\tan(90^\circ-\theta)=\cot\theta=\dfrac{1}{\tan\theta}$. That means angles that add up to $90^\circ$ are best friends—they turn into each other’s reciprocals!
3. So we can pair the terms so that one bracket has $1+\dfrac{1}{\tan\theta}$ and another has $1-\tan\theta$. That combination collapses nicely and saves loads of work.

Once you do the pairing and a bit of tidy arithmetic the big scary product shrinks down to a neat little number, which (spoiler-free) is one of the options 6, 8, 9, or 10.

Step 1 Bring every angle into the first quadrant

So the product becomes

Step 2 Form two complementary pairs

Because $65^\circ+25^\circ=90^\circ$ and $70^\circ+20^\circ=90^\circ$, write

Step 3 Use the algebraic identity

For any $t=\tan\theta$,

Apply it to each pair:

Step 4 Rewrite the whole product

Note that $80^\circ+10^\circ=90^\circ$, but $10^\circ$ is not present. Hence we switch to a direct numerical substitution for the two leftover brackets.

Step 5 Insert standard tangent values (4-figure table or calculator)

Compute each factor:

Now multiply:

Hence

$P=8\quad\text{(exact answer among the given options).}$

Option (B) 8 is correct.