Question

Prove that $\left( {{\text{1 + cotA - cosecA}}} \right)\left( {{\text{1 + tanA + secA}}} \right) = 2$

Answer 1

In order to solve this question easily we will transform the given terms in sin and cos. In this question we have to prove that the left-hand side is equal to the right-hand side by using some basic formula and identity as discussed below.

Complete step-by-step solution:
Now, by using trigonometric identities we will easily solve the given problem.
We know that
$\left( {{\text{1 + cotA - cosecA}}} \right)\left( {{\text{1 + tanA + secA}}} \right) = 2$
Now, first we consider left hand side of the given expression
LHS = $\left( {{\text{1 + cotA - cosecA}}} \right)\left( {{\text{1 + tanA + secA}}} \right)$
= $\left( {{\text{1 + cotA - cosecA}}} \right)\left( {{\text{1 + tanA + secA}}} \right)$
converting above expression n terms of sinA or cosA
= $\left( {{\text{1 + }}\dfrac{{{\text{cosA}}}}{{{\text{sinA}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{sinA}}}}} \right)\left( {{\text{1 + }}\dfrac{{{\text{sinA}}}}{{{\text{cosA}}}} + \dfrac{{\text{1}}}{{{\text{cosA}}}}} \right)$
= $\left( {{\text{1 + }}\dfrac{{{\text{cosA - 1}}}}{{{\text{sinA}}}}} \right)\left( {{\text{1 + }}\dfrac{{{\text{sinA + 1}}}}{{{\text{cosA}}}}} \right)$
Simplify above expressions as below-
= $\left( {\dfrac{{{\text{sinA + cosA - 1}}}}{{{\text{sinA}}}}} \right)\left( {\dfrac{{{\text{cosA + sinA + 1}}}}{{{\text{cosA}}}}} \right)$
= $\dfrac{{{{\left( {{\text{sinA + cosA}}} \right)}^{\text{2}}}{\text{ - 1}}}}{{{\text{sinAcosA}}}}$ As we know, $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right){\text{ = }}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)$
= $\dfrac{{{{\left( {{\text{sinA + cosA}}} \right)}^{\text{2}}}{\text{ - 1}}}}{{{\text{sinAcosA}}}}$ As we know, ${\left( {{\text{a + b}}} \right)^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{\text{ + 2ab}}$
= $\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{A + co}}{{\text{s}}^{\text{2}}}{\text{A + 2sinAcosA - 1}}}}{{{\text{sinAcosA}}}}$
As we know, ${\text{si}}{{\text{n}}^{\text{2}}}{\text{A + co}}{{\text{s}}^{\text{2}}}{\text{A = 1}}$
= $\dfrac{{{\text{1 + 2sinAcosA - 1}}}}{{{\text{sinAcosA}}}}$
= $\dfrac{{{\text{2sinAcosA}}}}{{{\text{sinAcosA}}}}$
= 2
Therefore, LHS=RHS.
Hence, proved $\left( {{\text{1 + cotA - cosecA}}} \right)\left( {{\text{1 + tanA + secA}}} \right) = 2$

Note: We need to remember some basic formulas related to trigonometry. So that we easily understand the problem and apply these formulas. Some of them are mentioned below which we used in this question.
These Identities are given as-
${\text{sin}}\theta$ = $\dfrac{{\text{1}}}{{{\text{cosec}}\theta }}$
${\text{cos}}\theta$ = $\dfrac{{\text{1}}}{{{\text{sec}}\theta }}$
${\text{tan}}\theta$ = $\dfrac{{\text{1}}}{{{\text{cot}}\theta }}$
${\text{cot}}\theta$= $\dfrac{{{\text{cos}}\theta }}{{{\text{sin}}\theta }}$